Prove that $\lim_{n \to \infty} \frac{n}{(n!)^\frac{1}{n}} = e $ I have solved the problem in just 2 lines using a theorem which asserts that
"Let ${u_n}$ be a real sequence such that $u_n > 0 \forall n \in \mathbb{N}$ and $\lim_{n \to \infty} \frac{u_{n+1}}{u_n} = \ell$ (finite of infinite). Then $\lim_{n \to \infty} (u_n)^{1 \over n} =\ell$ "
To prove the aforesaid limit, I fix $u_n={n^n \over n!}$. Then $u_n>0 \forall n \in \mathbb{N}$ and $\lim_{n \to \infty}\frac{u_{n+1}}{u_n}= \lim_{n \to \infty}(1+{1 \over n})^n=e$.
Then it follows from the above theorem that $\lim_{n \to \infty} (u_n)^{1 \over n} =e$  i.e. $ \lim_{n \to \infty} \frac{n}{(n!)^\frac{1}{n}} = e $. (Proved)
But I am trying to prove it without using the theorem. I am trying to get a generic proof.
Can anyone provide a proper wayout for this?
Thanks for your help in advance.
 A: EDIT: As pointed out in the comments, even though the final inequality is correct, it is insufficient since $(n+1)^{1/n} \to 1$ as $n \to \infty$. The lower bound can be obtained as shown in @adfriedman's answer.
Here's my take on it:
Whenever $n \geq 3$, we have
$$ n^n \geq (n+1)!, $$
and thus
$$ n^n \geq (n+1)n! \quad \Leftrightarrow \quad \frac{n}{n!^{\frac{1}{n}}} \geq (n+1)^{\frac{1}{n}}. $$
On the other hand, the Taylor expansion of $e^n$ gives
$$ \frac{n^n}{n!} \leq \sum_{k=0}^{\infty} \frac{n^k}{k!} = e^n \quad \Leftrightarrow \quad \frac{n}{n!^{\frac{1}{n}}} \leq e. $$
So,
$$ (n+1)^{\frac{1}{n}} \leq \frac{n}{n!^{\frac{1}{n}}} \leq e. $$
Apply the Squeeze Theorem.
A: Hint.
Use the Stirling asymptotic approximation formula
$$
n! = \sqrt{2\pi n}\left(\frac{n}{e}\right)^n(1+O(1/n))
$$
because
$$
\frac{n}{(n!)^{1/n}}= \frac{n}{(2\pi n)^{1/(2n)}(n/e)}(1+O(1/n)) = \frac{e}{(2\pi n)^{1/(2n)}}(1+O(1/n))
$$
hence 
$$
\lim_{n\rightarrow \infty}\frac{n}{(n!)^{1/n}} = e
$$
A: First, write
$$\ln\left(\frac{n}{(n!)^{1/n}}\right)
=\ln(n) - \tfrac{1}{n}\ln(n!) = \ln(n) - \tfrac1n \sum_{k=1}^n \ln(k)
= \ln(n) - \tfrac1n S_n$$
Consider upper and lower approximations of the integral of $\ln(x)$:
$$S_n = \sum_{k=1}^n \ln(k) \leq \int_1^{n+1}\ln(x)\;dx \leq \sum_{k=2}^{n+1}\ln(k) = S_n + \ln(n+1)$$
Hence
$$S_n  \leq \int_1^{n+1}\ln(x)\;dx = \left[ x\ln(x)-x \right]_1^{n+1} = (n+1)\ln(n+1) - n$$ and $$\int_1^{n+1}\ln(x)\;dx - \ln(n+1) = n\ln(n+1) - n \leq S_n$$
It follows:
$$\ln(n)-\tfrac1n \big((n+1)\ln(n+1) - n\big) \leq \ln(n) - \tfrac1n S_n \leq \ln(n) - \tfrac1n\big(n\ln(n+1) - n\big) $$
which simplifies to 
$$\ln\Bigg(\underbrace{\frac{n}{(n+1)^{1+\tfrac1n}}}_{\to \; 1}\Bigg)  + 1
\leq \ln(n)- \tfrac1n S_n \leq \ln\Bigg(\underbrace{\frac{n}{n+1}}_{\to\; 1}\Bigg) + 1$$
By the squeeze theorem $\ln\left(\frac{n}{(n!)^{1/n}}\right) = \ln(n) - \tfrac1n S_n \to 1$ as the log terms on both sides approach $\ln(1) = 0$, hence
$$\frac{n}{(n!)^{1/n}} = \exp\left(\ln\left(\frac{n}{(n!)^{1/n}}\right)\right) \to \exp(1) = e$$
