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.

i think you can do this with cayley-hamilton theorem which says that a matrix satisfies its characteristic equation. that is $$A^2 -(a+b)A + abI = 0  $$ or $$A^2 = (a+b)A - abI$$  we can use this to write every power of $A$ as a lineay combination of $A, I$. we will the division algorithm to find the remainder. suppose $$x^n = q(x)(x-a)(x-b) + \alpha x + \beta $$ putting $x = a, x = b$ gives $$\pmatrix{a^n\\b^n} = \pmatrix{a &1\\b&1}\pmatrix{\alpha\\\beta} \to \pmatrix{\alpha\\\beta}=\frac{1}{a-b}  \pmatrix{1 &-1\\-b&a}\pmatrix{a^n\\b^n}$$
and $$A^n = \frac{1}{a-b}\left(\left(a^n - b^n\right)A + \left(ab^n-a^nb\right)\right)= \frac{1}{a-b}\left(a^n\left(A- bI\right) +b^n \left(aI-A\right)\right).$$
