How to prove that $\lim\limits_{n \to \infty} \frac{k^n}{n!} = 0$ It recently came to my mind, how to prove that the factorial grows faster than the exponential, or that the linear grows faster than the logarithmic, etc...

I thought about writing:
$$
a(n) = \frac{k^n}{n!} = \frac{ k \times k \times \dots \times k}{1\times 2\times\dots\times n} = \frac k1 \times \frac k2 \times \dots \times \frac kn = \frac k1 \times \frac k2 \times \dots \times \frac kk \times \frac k{k+1} \times \dots \times \frac kn
$$
It's obvious that after k/k, every factor is smaller than 1, and by increasing n, k/n gets closer to 0, like if we had $\lim_{n \to \infty} (k/n) = 0$, for any constant $k$.

But, I think this is not a clear proof... so any hint is accepted.
Thank you for consideration.
The series for $e^k$
$$
\sum_{n=0}^\infty\frac{k^n}{n!}
$$
converges by the ratio test. The terms of a convergent series must tend to $0$.

For $n\ge2k$, the ratio of terms is $\frac{k^{n+1}/(n+1)!}{k^n/n!}=\frac{k}{n+1}&lt;\frac{1}{2}$. We can remove the reference to series (which seems to have bothered someone) with the following sandwich, valid for $n\ge2k$:
$$
0\le\frac{k^n}{n!}\le\frac{k^{2k}}{(2k)!}\left(\frac{1}{2}\right)^{n-2k}
$$
which shows that $\displaystyle\lim_{n\to\infty}\frac{k^n}{n!}=0$.
Let's observe an example :

$$\frac{9^n}{n!}$$ so we have that :

$$\frac{9^1}{1!}=9 , \frac{9^2}{2!}=40.5 , \frac{9^3}{3!}=121.5$$

but...

$$\frac{9^9}{9!}=\frac{9^8\cdot 9}{8! \cdot 9}=\frac{9^8}{8!}$$

and further...

$$\frac{9^{10}}{10!}=\frac{9^9\cdot 9}{9! \cdot 10} &lt; \frac{9^9}{9!}$$

So we may conclude: after we throw out the first $k-1$ elements that sequence starts to decrease  , therefore sequence :

$$\frac{k^n}{n!}$$

is decreasing if we set $ n\geq k$ , so we may conclude that :

$$n \rightarrow \infty \Rightarrow \frac{k^n}{n!} \rightarrow 0$$