Dividing $n^k+1$ by $n+1$ if and only if $k$ is odd For that question, I can use modular arithmetic to prove divisibility. Look at the following:
$$n \equiv-1\mod(n+1)$$
raising to $k^{th}$ power, if $k$ is odd, then
$$n^k \equiv(-1)^k \equiv-1\mod(n+1)$$ hence
$$n^k+1 \equiv0\mod(n+1)$$ as desired. On the other hand, if $k$ is odd, then
$$n^k \equiv(-1)^k \equiv1\mod(n+1)$$ hence
$$n^k+1 \equiv2\mod(n+1)$$
However, is it possible to come up with a divisibility proof? i.e. if $(n+1)k = (n^k+1)$, what is $k$??
I applied a long division but found $(n^{k-1}-n^{k-2}+n^{k-3}-n^{k-4}...)$ which I am suspicious about as I found in my instructor notes that if $k$ is odd, then $(n^{k}+1)=(n+1)(n^{k-1}-n^{k-2}...-n+1)$. This solution, the instructor's one, feels reasonable as when multiplying these terms, the result is $n^k+1$. What about mine?, how to connect it to the parity of $k$ as well?

$((n+1)-1)^k +1=$
$(n+1)^k + k(-1)(n+1)^{k-1}+......$
$..+k(n+1)(-1)^{k-1}+(-1)^k +1.$
All  terms, except the last term $(-1)^k$,  in the binomial expansion have a factor $(n+1)$.
For odd $k$: $(-1)^k +1=0.$
