"Suppose $5\mid k^5-k$ for some positive integer $k$. Then
\begin{align}
(k+1)^5-(k+1)
&= [(k+1)^5-k^5-1]+k^5-k \\
&= (5k^2+10k^3+5k^4)+k^5-k\\
&=5k^2(k+1)^2+k^5-k.
\end{align}
Since $5\mid 5k^2(k+1)^2$ and $5\mid k^5-k$, we conclude that $5 \mid (k+1)^5-k^5$. Since $5$ divides $1^5-1=0$, we conclude by induction on $k$ that $5\mid n^5-n$ is true for any positive integer $n$."
There are many integral representations for $\zeta(3)$
Some lesser known are for instance :
$$\int_0^1\frac{x(1-x)}{\sin\pi x}\text{d}x= 7\frac{\zeta(3)}{\pi^3} $$
$$\int_0^1 \frac{\operatorname{li}(x)^3 \space (x-1)}{x^3} \text{d}x = \frac{\zeta(3)}{4} $$
$$\int_0^\pi x(\pi - x) \csc(x) \s...
