If $k$ is composite, then $2^{k} -1$ is composite for $k \geq 2$ I am trying to prove that if $k$ is composite, then $2^{k} -1$ is composite, $k \geq 2$. 
I have already established the equality $$n^k - 1 = (n-1)(n^{k-1} + ... + 1) \tag{*}$$ If I let $n=2$ and $k=ab$, I don't really get anything useful. The hint says to let $n=2^a$ if $k=ab$, but this gives a $2^{a^2b}$ on the LHS of $*$...
Any help is appreciated
Update: I understand now...

$$2^k -1= 2^{ab} - 1 = (2^a)^b - 1$$
$$2^a = x$$
$$x^b-1 = (x-1)(x^{b-1} + ... + 1) = (2^a-1)((2^a)^{b-1} + ... + 1)$$
Clearly neither of these two factors are $1$, as $a>2>1$, so $2^k-1$ is composite
