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?

Using Stirling's formula, both are trivial.
