TITLE: $a_{n+1}=a_n^{n+1}+a_n+1$
QUESTION [6 upvotes]: Let $a_1>1$ and $$a_{n+1}=a_n^{n+1}+a_n+1, \: \forall n \geq 1$$
  Edit: Prove that there is a real number $x \neq 0$ such that $$\lim_{n \to \infty}\frac{a_n}{x^{n!}}=1$$

I think that $x=a_1$ works, and I tried to use the Squeeze Theorem to prove that. From $a_{n+1}>a_n^{n+1}>a_{n-1}^{n(n+1)}>\dots>a_1^{(n+1)!}$ we get that $1<\frac{a_{n+1}}{a_1^{(n+1)!}}$ and then I tried to find another bound for $a_{n+1}$ such that $\frac{a_{n+1}}{a_1^{(n+1)!}}<b_{n+1} \to 1$, but nothing seems to work.

REPLY [4 votes]: Let
$$x = \lim_{n\to\infty} a_n^{1/n!}.$$
We know from @frame95's answer that $x$ exists. Now set $\delta_n = \log a_n - n! \log x$ and notice that
$$ \frac{\delta_{n+1}}{(n+1)!} = \frac{\delta_n}{n!} + \frac{1}{(n+1)!}\log\left( 1 + \frac{1}{a_n^n} + \frac{1}{a_n^{n+1}} \right). $$
Since we know that $\delta_n/n! \to 0$, recursively applying the above relation and taking limit shows that
$$ \frac{\delta_n}{n!} = -\sum_{k = 0}^{\infty} \frac{1}{(n+k+1)!}\log\left( 1 + \frac{1}{a_{n+k}^{n+k}} + \frac{1}{a_{n+k}^{n+k+1}} \right) $$
Now by utilizing simple estimations $a_{n+k} \geq a_n^{k!} \geq a_1^{k!}$ and $\log(1+t) \leq t$ for $t \geq 0$, we obtain the following very crude bound
$$ \left| \delta_n \right| \leq \sum_{k=0}^{\infty} \frac{n!}{(n+k+1)!} \cdot 2 a_1^{-k!}. $$
Since each summand is bounded by $2a_1^{-k!}$ and $\sum_{k=0}^{\infty} 2a_1^{-k!} < \infty$, dominated convergence theorem tells that
$$ \limsup_{n\to\infty} \left| \delta_n \right| \leq \sum_{k=0}^{\infty} \lim_{n\to\infty} \frac{n!}{(n+k+1)!} \cdot 2 a_1^{-k!} = 0. $$
Therefore $\delta_n \to 0$ and hence $a_n!/x^{n!} = e^{\delta_n} \to 1$ as required.