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?

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.
