I found by chance that $$ f(n) = \sum_{k=1}^{\infty}{\frac{1}{(n+1)^{k-1}}}n^k $$ seems to be equal to $n \cdot (n+1)$. Proving this is out of my capabilities, but I would still love to see why that equivalence seems to hold.
So: is there a proof or intuitive explanation for why $f(n)=n \cdot (n+1)$?
$$\sum_{k=1}^\infty \dfrac{1}{(n+1)^{k-1}}n^k = (n+1)\sum_{k=1}^\infty \left(\dfrac{n}{n+1}\right)^k = (n+1)\left(\dfrac{1}{1-\tfrac{n}{n+1}}-1\right) = n(n+1)$$