For matrices $C, D$, show that $(CD)^{100} \neq C^{100} D^{100}$ The question is to prove this is false:

$(CD)^{100} = C^{100}\cdot D^{100}$, where $C$ and $D$ are matrices.

I looked through my textbook and could not find a proof for this. 

$$(CD)^{100}=\underbrace{(CD)(CD)\cdots(CD)}_{100\text{ times}}$$
This is only true in the case of matrices for which multiplication is commutative, $CD=DC$.
In such a case, you can use associativity and commutativity to write
$$\begin{align*}\underbrace{(CD)(CD)\cdots(CD)}_{100\text{ times}}&=C\underbrace{(DC)(DC)\cdots(DC)}_{99\text{ times}}D\\[1ex]
&=C\underbrace{(CD)(CD)\cdots(CD)}_{99\text{ times}}D\\[1ex]
&=C^2D\underbrace{(CD)\cdots(CD)}_{98\text{ times}}CD^2\end{align*}$$
and so on to arrive at $C^{100}D^{100}$. But multiplication is not commutative in the general case. All you need is a counterexample, like the choice of $C$ and $D$ that the other answer provides.
