I've found exercise with binomial coefficients in Kostrikin's book.
Proof that
$\sum_{i=1}^n{{r+1}\choose{i}}\left(1^i+2^i+\dots+n^i\right)=(n+1)^{r+1}-(n+1)$
I was trying to check that for small integers like $r=2$ and $n=1,2$ but i think that there is something wrong. My results didn't match.
For $r=2$ and $n=1$ we have ${{3}\choose{1}}\left(1^1\right)=3\not=2^3-2=6$. For $r=2$ and $n=2$, ${{3}\choose{1}}\left(1^1\right)+{{3}\choose{2}}\left(1^2+2^2\right)=18\not=3^3-3=24$. I can't find this identity anywhere. Is this identity known? How can i proof that without mathematical induction?