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. 
 A: A matrix corresponds to some (linear) operation on a vector. Let's take 2-component vectors and actions $C$ "make first component zero" and $D$ "swap components". Then $(CD)^{100}$ is "swap components, then make first component zero, repeat $100$ times" (which gives you zero vector, obviously) while $C^{100}D^{100}$ is "swap components $100$ times, then make first component zero $100$ times" (which is the same as $C$).
Now, how to get a matrix by operation description? Easy: first column is the result of operation on vector $\pmatrix{1 \\ 0}$, second column is the result of operation on vector $\pmatrix{0 \\ 1}$. Since $C \pmatrix{1 \\ 0} = \pmatrix{0 \\ 0}$, $C \pmatrix{0 \\ 1} = \pmatrix{0 \\ 1}$, we have $C = \pmatrix{0 & 0 \\ 0 & 1}$; likewise, $D = \pmatrix{0 & 1 \\ 1 & 0}$. 
Easy to see that $D^{100} = \pmatrix{1 & 0 \\ 0 & 1}$, $C^{100} = C$, $CD = \pmatrix{0 & 0 \\ 1 & 0}$ and $(CD)^{100} = \pmatrix{0 & 0 \\ 0 & 0} \neq C^{100}D^{100}$.
A: $$(CD)^{100}=\underbrace{(CD)(CD)\cdots(CD)}_{100\text{ times}}$$
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.
