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?

For your division, what is the last term? The general term will be $(-1)^{r-1}n^{k-r}$ and you will end when $r=k$ with $(-1)^{k-1}n^0=(-1)^{k-1}$.
Here you get $+1$ when $k$ is odd and $-1$ when $k$ is even. All that remains is to compute the remainder.
One way of doing this is simply to use the remainder theorem - if $$p(n)=(n+1)q(n)+r$$ using polynomial division, then $p(-1)=r$, and apply it to the polynomial $p(n)=n^k+1$
Indeed, it is not necessary to compute $q(n)$ specifically to use this result.
