Let $A$ be a complex $2$ by $2$ matrix having distinct eigenvalues $a, b$. Show that $A^n =\frac{ a^n}{a - b}(A - bI) + \frac{b^n}{b - a}(A - aI)$. 
Let $A\in\mathscr{M}_{2\times 2}(\mathbb{C})$ be a matrix having distinct eigenvalues $a\neq b$. Show that, for all $n > 0$,
  \begin{equation*}
A^n =\frac{ a^n}{a - b}(A - bI) + \frac{b^n}{b - a}(A - aI).
\end{equation*} (Exercise 707 from Golan, The Linear Algebra a Beginning Graduate Student Ought to Know.)

I proved it by induction, under the assumption that it is true for $A,A^2$,
\begin{equation*}
A^n =\frac{ a^n}{a - b}(A - bI) + \frac{b^n}{b - a}(A - aI)=\frac{(a^n-b^n)A+(ab^n-a^nb)I}{a-b}\\\Rightarrow A^{n+1}=A\frac{(a^n-b^n)A+(ab^n-a^nb)I}{a-b}=\frac{(a^n-b^n)A^2+(ab^n-a^nb)A}{a-b}\cdots
\end{equation*}
but I can't prove it for $A^2$.
I'm also wondering if there is another way to prove it by induction just starting with $A$ or a straight way.
Thanks.

We have
$$A = X \begin{bmatrix}a & 0\\0 & b\end{bmatrix} X^{-1}$$
This means
$$A^n = X \begin{bmatrix}a^n & 0\\0 & b^n\end{bmatrix} X^{-1}$$
We have
$$A-bI = X \begin{bmatrix}a-b & 0\\0 & 0\end{bmatrix} X^{-1} \text{ and }A-aI = X \begin{bmatrix}0 & 0\\0 & b-a\end{bmatrix} X^{-1}$$
Hence,
$$\dfrac{a^n}{a-b}\left(A-bI\right) = X \begin{bmatrix}a^n & 0\\0 & 0\end{bmatrix} X^{-1} \text{ and }\dfrac{b^n}{b-a}\left(A-aI\right) = X \begin{bmatrix}0 & 0\\0 & b^n\end{bmatrix} X^{-1}$$
Hence,
\begin{align}
\dfrac{a^n}{a-b}\left(A-bI\right) + \dfrac{b^n}{b-a}\left(A-aI\right) & = X \begin{bmatrix}a^n & 0\\0 & 0\end{bmatrix} X^{-1} + X \begin{bmatrix}0 & 0\\0 & b^n\end{bmatrix} X^{-1}\\
& = X \begin{bmatrix}a^n & 0\\0 & b^n\end{bmatrix} X^{-1} = A^n
\end{align}
