Mathematical Induction Matrix Example I'm a little rusty and I've never done a mathematical induction problem with matrices so I'm needing a little help in setting this problem up. 
Show that $$\begin{bmatrix}1&1\\1&1\end{bmatrix}^{n} = \begin{bmatrix}2^{(n-1)}&2^{(n-1)}\\2^{(n-1)}&2^{(n-1)}\end{bmatrix}$$  for every $n\ge 1$.
 A: The case $n=1$ is clear since $2^0 = 1$. So suppose that
$$\begin{pmatrix} 1&1 \\ 1 & 1 \end{pmatrix}^n = \begin{pmatrix} 2^{n-1}&2^{n-1} \\ 2^{n-1} & 2^{n-1}\end{pmatrix} \quad \quad *$$
for some $n \geq 1$ and let us prove that
$$\begin{pmatrix} 1&1 \\ 1 & 1 \end{pmatrix}^{n+1} = \begin{pmatrix} 2^{n}&2^{n} \\ 2^{n} & 2^{n}\end{pmatrix}. $$
We have
$$\begin{pmatrix} 1&1 \\ 1 & 1 \end{pmatrix}^{n+1} = \begin{pmatrix} 1&1 \\ 1 & 1 \end{pmatrix}^{n} \begin{pmatrix} 1&1 \\ 1 & 1 \end{pmatrix} \overset{*}{=} \begin{pmatrix} 2^{n-1}&2^{n-1} \\ 2^{n-1} & 2^{n-1}\end{pmatrix}\begin{pmatrix} 1&1 \\ 1 & 1 \end{pmatrix}  = \begin{pmatrix} 2\cdot 2^{n-1}&2\cdot 2^{n-1} \\ 2\cdot 2^{n-1} & 2\cdot 2^{n-1}\end{pmatrix} =\begin{pmatrix} 2^{n}&2^{n} \\ 2^{n} & 2^{n}\end{pmatrix}. $$
And thus the relation is true for every $n \in \mathbb{N}$
A: It is clearly true for $n=1$. Assume it's true for $n$. Then 
$$\begin{pmatrix}1&1\\1&1\end{pmatrix}^{n+1}=\begin{pmatrix}1&1\\1&1\end{pmatrix}\begin{pmatrix}2^{(n-1)}&2^{(n-1)}\\2^{(n-1)}&2^{(n-1)}\end{pmatrix}=\begin{pmatrix}2^{(n-1)}+2^{(n-1)}&2^{(n-1)}+2^{(n-1)}\\2^{(n-1)}+2^{(n-1)}&2^{(n-1)}+2^{(n-1)}\end{pmatrix}=\begin{pmatrix}2^n&2^n\\2^n&2^n\end{pmatrix}$$
So it's true for $n+1$. By induction it is true for all $n$.
