(The presentation that follows is a very light introduction. For more austere results, consult Manuscript 1).

Let's begin with some notation. Let n in N be a positive integer. We define as usual:

n! = 1*2*3*...*(n-1)*n.

Now we will define a new sequence of iterated factorials for n in N, as follows:

hf(n,k) = {n! iff k=1, [hf(n,k-1)]!, iff k > 1}. (1)

hf(n,2) = [n!]!, hf(n,5) = [[[[n!]!]!]!]!, etc. Where no confusion shall arise, we will use hf(n,k) = n!!...!, (k-"!"s). Some examples with Maple:

3!! = 720

4!! = 620448401733239439360000

3!!! =~ .2601218944 * 10^{1747}

etc.

We also use the hyperpower function defined here as:

F(x,y,k) = {x^{y} iff k =1, x^{F(x,k-1)}, iff k > 1}, so,

F(x,1,1) = x,

F(x,1,2) = ^{2}x = x^{x},

F(x,y,2) = x^{xy}

...

F(x,1,k) = ^{k}x = x^{xx...x}
(k-x's).

First, some elementary facts.

The comparison property of divisibility:

a | b and b =/= 0, => |a| ≤ |b|. (1)

Proof:

In any standard Number Theory book.

Corollary:

a | b and a, b > 0, => a≤b.

A couple of immediate facts about the sequence hf(n,k):

Lemma #1:

If m and n are in N, then: m≤n <=> hf(m,k) ≤ hf(n,k), for all positive k.

Proof:

Trivial by induction on k. The factorial sequence is 1-1 and strictly increasing.

Lemma #2:

If m and n are in N, then m≤n <=> m! | n!

Proof:

If m≤n, clearly m! | n!. On the other hand, m! | n!, along with the corollary, => m! ≤n!, and the lemma follows from lemma #1 (for k=1).

Lemma #3:

If m and n are in N, then: m≤n <=> hf(m,k) | hf(n,k), for all k in N.

Proof:

By induction on k. For k=1, this is lemma #2. Assume that the lemma holds for k in N, k>1. That is, m≤n <=> hf(m, k) | hf(n,k). From the induction hypothesis, m≤n, => hf(m,k) | hf(n,k). From the corollary, => hf(m,k) ≤ hf(n,k). From lemma #2, => [hf(m,k)]! | [hf(n,k)]!, => hf(m,k+1) | hf(n, k+1).

For the reverse, assume hf(m,k+1) | hf(n, k+1). From the corollary, => hf(m,k+1) ≤ hf(n,k+1). From lemma #1 (for arbitrary k), => m≤n, and the lemma follows.

Let's now look how fast these hyperfactorials grow. We will need the following two lemmas along with their corollaries.

Lemma #4:

If k ≥ 1 then for all n in N, hf(n+1,k+1) - hf(n,k+1) >
hf(n,k+1)*[n^{hf(n+1,k) - hf(n,k)} - 1]

Proof:

Directly or by induction. The direct proof is shorter. By lemma #3, hf(n,k+1) | hf(n+1,k+1), => hf(n+1,k+1) - hf(n,k+1) = 1*2*...*hf(n+1,k) - 1*2*...*hf(n,k) = 1*2*...*hf(n,k)*[(hf(n,k)+1)*(hf(n,k)+2)*...(hf(n+1,k)) - 1] = hf(n,k+1)*[(hf(n,k)+1)*(hf(n,k)+2)*...(hf(n+1,k)) - 1]. If k ≥1, => hf(n,k) + j > n, for all j. There are hf(n+1,k) - hf(n,k) terms in the product, and the lemma follows.

Corollary:

If k ≥1 then for all n in N, hf(n+1,k) - hf(n,k) > ^{k}n =
F(n,1,k)

Sketch of Proof:

Iterate the weaker inequality of lemma #4. hf(n+1,k+1) - hf(n,k+1) >
n^{hf(n+1,k) - hf(n,k)}, and use induction.

Lemma #5:

If k ≥1, then hf(n,k)/hf(n+1,k) < 1/^{k}n = 1/F(n,1,k)

Proof:

Using lemma #3, hf(n,k) | hf(n+1,k), => hf(n,k)/hf(n+1,k) =

Using the corollary above, the product contains at most ^{k-1}n terms, and
hf(n,k-1) + j > n, for all j, => hf(n,k)/hf(n+1,k) < 1/n
^{k-1n} = 1/^{k}n and the lemma follows.

Lemma #6:

If k ≥1, then

converges for all x.

Proof:

If k=1, we get the exp series. If k>1 then |a(n+1)/a(n)| = |x*hf(n,k)/hf(n+1,k)|. By lemma #5, the above tends to 0, for all x.

Lemma #7:

If k =1 then the series,

converges for |x| < 1/e. If k > 1 then it converges for all x.

Proof:

If k=1, the Ratio test gives lim_{n->+∞}|a(n+1)/a(n)| = |ex| <
1, provided |x| < 1/e. >If k > 1, the same test gives |a(n+1)/a(n)| =
(n+1)^{(n+1)}*x/[n^{n}*(n+1)] *(n+1)*hf(n,k)/hf(n+1,k)|. Using lemma
#5, |a(n+1)/a(n)| < |(n+1)^{(n+1)}*x/[n^{n}*(n+1)]
*(n+1)/^{k}n|

lim_{n->+∞}[(n+1)^{(n+1)}*x/[n^{n}*(n+1)]] = ex, as
before, and obviously lim_{n->+∞}(n+1)/^{k}n = 0, for k >
1, and the rest of the lemma follows.

Can we up the stakes a bit? What about x^{xn} on the numerator of
the series instead of x^{n}? Let's see.

Lemma #8:

If k > 2 then the series:

converges for all x.

Proof:

Using lemma #5 |a(n+1)/a(n)| < |x^{xn*[x-1]}/^{k}n|.
For sufficiently large n, x < n, => |a(n+1)/a(n)| <
|n^{nn}/^{k}n|=|^{3}n/^{k}n|, so if k >
2, => lim_{n->+∞}|a(n+1)/a(n)| < 1, and the lemma follows.

Lemma #9:

If k > m - 1 then the series,

converges for all x.

Proof:

Similar to the proof of lemma #8.

Let's see if we can go further up. What about ^{n}x on the numerator?

Lemma #10:

If k ≥1 then the series,

converges for x in (0,e^{(1/e)}].

Proof:

Using lemma #5, |a(n+1)/a(n)| < |^{n+1}x/^{n}x *
1/^{k}n|. When x is in [(1/e)^{e},e^{(1/e)}],
lim_{n->+∞}^{n}x = e^{-W(-ln(x))}, by the article on
Infinite Exponentials, so,
lim_{n->+∞}^{n+1}x/^{n}x = 1, while
lim_{n->+∞}1/^{k}n = 0, for k ≥1, and the lemma
follows.

If on the other hand, x is in (0,(1/e)^{e}), |a(n+1)/a(n)| =
|^{n+1}x/^{n}x*hf(n,k)/hf(n+1,k)|, and ^{n}x is a two cycle, bounded above by 1 and below by 0. This means that we
may have problems with the odd subsequence: ^{2m+1}x, on the denominator. Using
lemma #5, => |a(n+1)/a(n)| < |^{n+1}x/^{n}x*1/^{k}n|, as
before, and lim_{m->+∞}^{2m}x = a < 1, with a being a
solution to x^{xa}=a, (See Solving the Second Real Auxiliary Exponential
Equation), lim_{m->+∞}^{2m+1}x = b > 0, b < a, and
with b given by x^{a} = b, above. This means that
|^{n+1}x/^{n}x|, n in N, is a two-cycle. We check the even and odd
subsequences: lim_{m->+∞}|a(2m+1)/a(2m)| ≤
lim_{m->+∞}|^{2m+1}x/^{2m}x*1/^{k}(2m)| =
|b/a*0| = 0, while, lim_{m->+∞}|a(2m+2)/a(2m+1)| ≤
lim_{m->+∞}|^{2m+2}x/^{2m+1}x*1/^{k}(2m+1)| =
|a/b*0| = 0, and the lemma follows^{[1]}.

Lemma #11

If k >1 then for all n in N, hf(n+1,k+1) - hf(n,k) > hf(n,k)*[n^{hf(n+1,k)
- hf(n,k-1)} - 1]

Proof:

Directly or by induction. The direct proof is shorter. By lemma #3, hf(n,k) | hf(n+1,k), and obviously hf(n+1,k) | hf(n+1,k+1), => hf(n+1,k+1) - hf(n,k+1) = 1*2*...*hf(n+1,k) - 1*2*...*hf(n,k-1) = 1*2*...*hf(n,k-1)*[(hf(n,k-1)+1)*(hf(n,k-1)+2)*...(hf(n+1,k)) - 1] = hf(n,k)*[(hf(n,k-1)+1)*(hf(n,k-1)+2)*...(hf(n+1,k)) - 1]. If k >1, => hf(n,k-1) + j > n, for all j. There are hf(n+1,k) - hf(n,k-1) terms in the product, and the lemma follows.

Corollary:

If k >1 then for all n in N hf(n+1,k+1) - hf(n,k) > ^{k+1}n =
F(n,1,k+1)

Proof

Iterate the weaker inequality of lemma #11, hf(n+1,k+1) - hf(n,k) >
n^{hf(n+1,k) - hf(n,k-1)}, and use induction.

Lemma #12:

If k >1, then, hf(n,k)/hf(n+1,k+1) < 1/^{k+1}n = 1/F(n,1,k+1)

Proof:

By lemma #3, hf(n,k) | hf(n+1,k), and obviously hf(n+1,k) | hf(n+1,k+1), => hf(n,k)/hf(n+1,k+1) =

Using the corollary above, the product contains at most ^{k}n terms, and
hf(n,k-1) + j > n, for all j, => hf(n,k)/hf(n+1,k)< 1/n
^{kn} = 1/^{k+1}n and the lemma follows.

We can still go further.

Lemma #13:

The series,

converges for all x > 0.

Proof:

Using lemma #12, |a(n+1)/a(n)| <
|^{n+1}x/^{n}x*1/^{n+1}n|. If x is in
[(1/e)^{e},e^{(1/e)}] then lim_{n->+∞}^{n}x =
e^{-W(-ln(x))}, so, =>
lim_{n->+∞}|^{n+1}x/^{n}x| = 1, as before and
lim_{n->+∞}|1/^{n+1}n| = 0, and the result follows. If x >
e^{(1/e)}, for sufficiently large n, x < n, => ^{n+1}x <
^{n+1}n. Using lemma #12, => |a(n+1)/a(n)| < |1/^{n}x|,
consequently, lim_{n->+∞}|a(n+1)/a(n)| = 0, and the result
follows.

If on the other hand, x is in (0,(1/e)^{e}), |a(n+1)/a(n)| =
|^{n+1}x/^{n}x*hf(n,n)/hf(n+1,n+1)|, and ^{n}x is again a
two cycle, bounded above by 1 and below by 0. Using lemma
#12, |a(n+1)/a(n)| < |^{n+1}x/^{n}x*1/^{n+1}n|, as before,
and lim_{m->+∞}^{2m}x = a < 1, with a being a solution to
x^{xa}=a, (See Solving the
Second Real Auxiliary Exponential Equation),
lim_{m->+∞}^{2m+1}x = b > 0, b < a, and with b given by
x^{a} = b, above. This means that, |^{n+1}x/^{n}x|, n in N, is
a two-cycle. We check the even and odd subsequences:
lim_{m->+∞}|a(2m+1)/a(2m)| ≤
lim_{m->+∞}|^{2m+1}x/^{2m}x*1/^{2m+1}(2m)| =
|b/a*0| = 0, while, lim_{m->+∞}|a(2m+2)/a(2m+1)| ≤
lim_{m->+∞}|^{2m+2}x/^{2m+1}x*1/^{2m+2}(2m+1)|
= |a/b*0| = 0, and the lemma follows.

- It is non-trivial to see what happens for x in (e
^{(1/e)},+∞). Try to investigate yourself what the relation between k and x should be, so that lim_{n->+∞}|^{n+1}x/^{n}x*hf(n,k)/hf(n+1,k)| is 0, a > 0, or +∞.