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...

We have in general
$$ 2^n-1 = \underset {n \text { times}} { \underbrace {1 + 2^1 + 2^2 + ... + 2^{n-1}}}$$
If $n$ is composite, say $a \cdot b$ then
$$\begin{array}{lll} 2^n-1 &= \underset { a \cdot b \text { times}} { \underbrace {1 + 2^1 + 2^2 + ... + 2^{n-1}}} \\ \phantom{X} \\
&=\underset {a \text{ times}} { \underbrace {
      \underset {b \text{ times}}{\underbrace{1+2^1+2^2+...+2^{b-1}}}
+ 2^b \underset {b \text{ times}}{\underbrace{1+2^1+2^2+...+2^{b-1}}}
+ ... 
+ 2^{b(a-1)}\underset {b \text{times}}{\underbrace{1+2^1+2^2+...+2^{b-1}}}
}} \\ \phantom{X} \\
&=(2^b-1) \cdot(1+2^b+2^{2b} + ... + 2^{(a-1)b})\\ \phantom{X} \\
&=(2^b-1) \cdot \left({ 2^{ab}-1 \over 2^b-1} \right) \end{array}$$
... is composite too (if $b>1$).
