$a^n+1$ prime $\Rightarrow n = 2^k$ is a power of $2$ I'm trying to prove that $a^n+1$ can only be prime if $n$ is a power of $2$. Is there a general factorization of $a^n+1$?

At the core it is the trivial identity $\rm\, (-1)^m \equiv -1\ $ for odd $\rm\,m.\,$ So if $\rm\,n\,$ has odd factor $\rm\,m,\ \ n = k m,\,$ 
then we have $\rm\ \ a^k\equiv -1\ \,\Rightarrow\,\ a^n =\, (a^k)^m\equiv -1\ \ (mod\ a^k+1),\ $ hence $\rm\ a^k+1\ $ divides $\rm\ a^n+1\,.$ 
Or $ $ put $\rm\ x = a^k\ $ in $\rm\ x+1\mid  x^m + 1,\, $ by the Factor Theorem: $\rm\ x-c\, \mid\, f(x)-f(c)\ $ in $\rm\ \mathbb Z[x].$ 
Note the this is simply the case $\rm\,m\,$ odd, $\rm\ x\to -x\ $ in this prior question today.
