I have no idea how to go about it really. Any help is very much appreciated.
$\displaystyle
\sum_{k=1}^{n}k^2 = \frac{n(n+1)(2n+1)}{6}
$
I already know about that, but thanks for the suggestion anyway. I am going to write here, what I have so far, and hopefully you or someone will point me in the right direction.
Proving for n = 1, easy.
Assuming it's true for n, let's evaluate for n+1.
$\displaystyle \sum_{k=1}^{n+1} k^{n+1} = \sum_{k=1}^n k^n + k^{n+1} = \frac{n(n+1)(2n+1)}{6} + k^{n+1}$
$\displaystyle \sum_{k=1}^{n+1} k^{n+1} = \frac { (n+1)(n+2)(2n+3)}{6}$
$\displaystyle \Rightarrow \frac{n(n+1)(2n+1)}{6} + k^{n+1} = \frac { (n+1)(n+2)(2n+3)}{6}$
$\displaystyle \Leftrightarrow n(n+1)(2n+1) + 6k^{n+1} = (n+1)(n+2)(2n+3)$
And I don't know how to go from here... I don't even know if what I've done is relevant...
No, not $\displaystyle k^{n+1}$! Your sum was $\displaystyle \sum_{k=1}^n k^2$. You want $\displaystyle \sum_{k=1}^{n+1} k^2= \sum_{k=1}^n k^2+ (k+1)^2$.
$\displaystyle \sum_{k=1}^{n+1} k^{n+1} = \frac { (n+1)(n+2)(2n+3)}{6}$
$\displaystyle \Rightarrow \frac{n(n+1)(2n+1)}{6} + k^{n+1} = \frac { (n+1)(n+2)(2n+3)}{6}$
$\displaystyle \Leftrightarrow n(n+1)(2n+1) + 6k^{n+1} = (n+1)(n+2)(2n+3)$
And I don't know how to go from here... I don't even know if what I've done is relevant...