Limit of $\sqrt[x-1]{\frac{x^x}{x!}}$ I'm trying to solve some task and I'm stuck. I suppose that I will be able to solve my problem, if I'll find elementary way to calculate $\lim_{x \to \infty}\sqrt[x-1]{\frac{x^x}{x!}}$ for $x \in \mathbb{N}_+$.
My effort: I had prove, that $x! \geq (\frac{x+1}{e})^x$, so (cause $x^x>x!$):
$$
\left(\frac{x^x}{x!}\right)^{\frac 1 x} \leq \left(\frac{x^x}{(x+1)^x}\right)^{\frac 1 x} \cdot e \xrightarrow{x \to \infty} e
$$
But how can I end that proof?
I will be grateful for all the advice.
 A: Say $y=\sqrt[x-1]{\frac{x^x}{x!}}$ 
Then, $\ln y=\frac{\ln(\frac{x^x}{x!})}{x-1}=-\frac{x}{x-1} \cdot \frac{\ln(\frac{x!}{x^x})}{x} =-\frac{x}{x-1} \cdot \sum \limits_{i=1}^{x} \ln{\frac{i}{x}}\cdot\frac{1}{x} \xrightarrow{x \to \infty} (-1) \cdot \int \limits_{0}^{1}\ln x ~dx=(-1) \cdot(-1)=1$
But, $\lim \limits_{x \to \infty} \ln y = \ln \left(\lim \limits_{x \to \infty}y\right)$.
So, $ \lim \limits_{x \to \infty}\sqrt[x-1]{\frac{x^x}{x!}} = e$
A: Elementary solution to this problem:
Fact:
$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = g \in \mathbb{R} \Longrightarrow \lim_{n\to\infty}\sqrt[n]{a_n} = g$$
So, we can take $a_n = \frac{n^n}{n!}$, then $\frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{n^n}\cdot\frac{n!}{(n+1)!}=(1+\frac{1}{n})^n (n+1) \cdot \frac{1}{(n+1)} = (1+\frac{1}{n})^n$
$$
\lim_{n\to\infty}\frac{a_{n+1}}{a_{n}} = \lim_{n \to \infty} (1+\frac{1}{n})^n = e \Longrightarrow
\lim_{n \to \infty}\sqrt[n]{a_n} = e
$$
Now just arithmetic properties of limits to calculate $\lim_{n\to\infty}\sqrt[n-1]{a_n}$. It's easy.

Above fact is easy to prove. Start with prove of lemma, let $(a_n)$ be sequence of positive numbers then:
$$\lim_{n\to\infty} a_n = g \in \mathbb{R} \Longrightarrow \lim_{n\to\infty}\sqrt[n]{\prod_{i=1}^{n} a_i} = g$$
Use logarithms. Then $\ln\sqrt[n]{\prod_{i=1}^{n}a_i} = \frac{1}{n}\sum \ln a_i$ and it's arithmetic average, from Stolz–Cesàro theorem we conclude then arithmetic average of first $n$ elements of $(a_n)$ goes to $g$, so (as $a_n \rightarrow g$)
$\frac{1}{n}\sum \ln a_i = \lim_{n\to\infty}\ln a_n \Rightarrow
\lim_{n\to\infty}\sqrt[n]{\prod_{i=1}^{n} a_i} = g
$. So lemma is correct.
Now take sequence $b_n = \frac{a_n}{a_{n-1}} \wedge b_1 = a_n$ and use lemma.
