Show that $\lim_{n\to\infty}\frac{a^n}{n!}=0$ and that $\sqrt[n]{n!}$ diverges. 

Let $a\in\mathbb{R}$. Show that
    $$
\lim_{n\to\infty}\frac{a^n}{n!}=0.
$$
    Then use this result to prove that $(b_n)_{n\in\mathbb{N}}$ with
    $$
b_n:=\sqrt[n]{n!}
$$
    diverges.


Okay, I think that's not too bad.
I write
$$
\frac{a^n}{n!}=\frac{a}{n}\cdot\frac{a}{n-1}\cdot\frac{a}{n-2}\cdot\ldots\cdot a
$$
and because all the factors converges to 0 resp. to $a$ (i.e. the limits exist) I can write
$$
\lim_{n\to\infty}\frac{a^n}{n!}=\lim_{n\to\infty}\frac{a}{n}\cdot\lim_{n\to\infty}\frac{a}{n-1}\cdot\ldots\cdot\lim_{n\to\infty}a=0\cdot 0\cdot\ldots\cdot a=0.
$$
Let $a_n:=\frac{a^n}{n!}$ and $a=1$ then
$$
b_n=\frac{1}{\sqrt[n]{a_n}}.
$$
Because (as shown above) $a_n\to 0$ it follows that $\sqrt[n]{a_n}\to 0$, because 
$$
\lvert\sqrt[n]{a_n}\rvert\leqslant\lvert a_n\rvert\to 0\implies\lvert\sqrt[n]{a_n}\rvert\to 0
$$
and therefore $b_n\to\infty$.

I think that's all. Am I right?
 A: For the first:
$\lim_{n \to \infty} |\frac{a^n}{n!}|\leq \lim_{n \to \infty} \frac{m^n}{n!}\leq\lim_{n \to \infty} \frac{m^n}{(n-m)!}=\lim_{n \to \infty} \frac{m\cdot m\cdot m\cdot m...}{ (m+1)\cdot(m+2)\cdot (m+3)\cdot (m+4)...}=0$ (n factors)
Where m is (in terms of absolute value) a bigger natural number than a.
For the second:
Its easy so see that 
$n!\geq(\frac{n}{2})^\frac{n}{2}$ (How?)
Now you obtain that:
$\sqrt[n]{n!}\geq \sqrt[n]{\frac{n}{2}^\frac{n}{2}}=\sqrt(\frac{n}{2}) \rightarrow \infty$
A: Your proof is not correct, because
$$|\sqrt[n]{a_n}| \leq \lvert a_n\rvert$$ is not true for $|a_n| < 1$ (in fact holds $\lvert\sqrt[n]{a_n}\rvert >\lvert a_n\rvert$).
I give you a hint. Suppose by contradiction that $b_n$ is not divergent (hence it is bounded), and call $a = \sup_{n\geq1} \sqrt[n]{n!}$. 
Then for every $n \geq 1$ you have $a^n \geq n!$.
But now, what can you say about the sequence $a_n=\frac{a^n}{n!}$? Can it be convergent to $0$?
A: For the first limit, hint: Let $a$ be any real number. As $n$ tends to $+\infty,$
$$
\left|\dfrac{\dfrac{a^{n+1}}{(n+1)!}}{\dfrac{a^n}{n!}}\right|=\left|\dfrac{a}{n+1}\right| < \frac{1}2, \quad n\geq2|a|,
$$ and, by an elementary recursion,
$$
\left|\dfrac{a^n}{n!}\right|< \frac{C(a)}{2^n}, 
$$ as $n$ is great, where $C(a)$ is a constant in the variable $n$, and your sequence then tends to zero.
For the second limit, you may apply Riemman integral, as $n$ tends to $+\infty,$
$$\large \sqrt[n]{n!}=\displaystyle  e^{\frac1n \sum_1^n\ln k}=e^{\frac1n \sum_1^n\ln (k/n)+\ln n}\sim n \: e^{\:\Large \int_0^1\ln t \:{\rm d}t }
$$ giving a  divergent sequence.
A: Using Stirling's formula, both are trivial.
A: If we have to conclude the second part from (the proof of) the first part, then here is one way.
Note that $a_{n+1}=\frac{a}{n+1}a_n$. So for all $\epsilon >0$ we have ${a\over n+1}<\epsilon$ for all large enough $n$. So there exists $k\in\mathbb{N}$ such that $0\leq |a_{n}|\leq \epsilon^{n-k}|a_k|$ for all $n\geq k$. Fixing $\epsilon<1$, since $k$ is fixed we conclude by the squeeze theorem that $a_n\rightarrow 0$.
Now for $a\neq 0$ we further have $\sqrt[n]{|a_n|}={|a|\over\sqrt[n]{n!}}\leq \epsilon^{1-{k\over n}} \sqrt[n]{|a_k|}\rightarrow \epsilon $ for all $\epsilon>0$. From this we conclude that $\sqrt[n]{n!}\rightarrow \infty$ as desired.
A: I think there is another way to prove the first part, simply using that
$$
\exp(a)=\sum_{n=0}^{\infty}\frac{a^n}{n!}
$$
and this this series converges for all $a\in\mathbb{R}$, i.e. that the summands are a null sequence. That's it.
For the second part: Suppose $(b_n)_n$ does converge. Then this sequence is bounded, let's say $\lvert b_n\rvert < M$. Then by what is shown we have
$$
\frac{b_n^n}{n!}<\frac{M^n}{n!}\to 0\implies \frac{b_n^n}{n!}\to 0,
$$
but $b_n^n/n!=1$.
