Proof $e^n*n!$ is an asymptote of $(n+1)^n$ I would like to prove $\lim_{n\to \infty}e^nn!-(n+1)^n=0$.
All I have really done is show $(n+1)^n=\sum_{i=0}^n\frac{n!}{(n+1)^i(i!)(n-i)!}$
 A: Stirling's formula seems to hold the answer.
$$ \frac{\alpha^n n!}{(n+1)^n} \sim \frac{\alpha^n\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{n^n}\sim \left(\frac{\alpha}{e}\right)^n\sqrt{n} $$
Since the exponent grows/shrinks quicker than $\sqrt{n}$ then if $\alpha\geq e$ it will diverge to $+\infty$ and if $\alpha<e$ the ratio will go to $0$...
A: (This is too long for a comment, but neither is it yet a full answer.)
Consider the Taylor series expansion of $e^x$.
\begin{align*}
e^x &= \sum_{k = 0}^\infty \frac{x^k}{k!}\\
&= \lim_{n \rightarrow \infty} \sum_{k = 0}^n \frac{x^k}{k!}
\end{align*}
It follows
\begin{align*}
\lim_{n \rightarrow \infty} n! e^n &= \lim_{n \rightarrow \infty} n! \sum_{k = 0}^n \frac{n^k}{k!}\\
&= \lim_{n \rightarrow \infty} \sum_{k = 0}^n (n-k)!\binom{n}{k} n^k.
\end{align*}
On the other hand, the Binomial Theorem gives
\begin{align*}
\lim_{n \rightarrow \infty} (1+n)^n &= \lim_{n \rightarrow \infty} \sum_{k=0}^n \binom{n}{k}n^k.
\end{align*}
(At this point, I'm not quite sure what to say about that factor of $(n-k)!$ except that it is least significant for the largest values of $k$.)
