Fermat primes relation to $2^n+1$ If $2^n+1$ is prime how to show that  $n$ must be a power of $2$
Being at elementary level I am at a loss what to begin with?

Suppose that $n$ is odd. Then $$x^n=(x+1-1)^n=(-1)^n=  -1\mod x+1$$ so that $x+1\mid x^n+1$. Thus if $2^n+1$ is not composite, it must be the case $n=2^u$. If you want to be explicit $$\begin{align}
  \frac{{{x^n} + 1}}{{x + 1}} &= \frac{{{{( - x)}^n} - 1}}{{( - x) - 1}} \cr 
   &= \frac{{{u^n} - 1}}{{u - 1}} \cr 
   &= 1 + u + {u^2} +  \cdots  + {u^{n - 1}} \cr 
   &= 1 - x + {x^2} -  +  \cdots  + {\left( { - 1} \right)^{n - 1}}{x^{n - 1}} \end{align} $$
