Matrix induction proof Given the following
$\lambda_{1}=\frac{1-\sqrt{5}}{2}$ and $\lambda_{2}=\frac{1+\sqrt{5}}{2}$
How do I prove this using induction:
$\begin{align*}
A^k=\frac{1}{\sqrt{5}}\left(\begin{array}{cc}
\lambda_2^{k-1}-\lambda_1^{k-1} & \lambda_2^{k}-\lambda_1^{k}\\
\lambda_2^{k}-\lambda_1^{k} & \lambda_2^{k+1}-\lambda_1^{k+1}
\end{array}\right),\,k>0
\end{align*}$
when $
A=\left(\begin{array}{cc}
0 & 1\\
1 & 1
\end{array}\right)\in\text{Mat}_{2}(\mathbb{R})
$.
I know how induction works and it holds for $k=1$. But I'm stuck at $k+1$.
My own work (Revision):
\begin{align*}
A^{k+1}=\frac{1}{\sqrt{5}}\left(\begin{array}{cc}
\lambda_2^{k-1}-\lambda_1^{k-1} & \lambda_2^{k}-\lambda_1^{k}\\
\lambda_2^{k}-\lambda_1^{k} & \lambda_2^{k+1}-\lambda_1^{k+1}
\end{array}\right)\cdot\left(\begin{array}{cc}
0 & 1\\
1 & 1
\end{array}\right)=\frac{1}{\sqrt{5}}\left(\begin{array}{cc}
\lambda_2^{k}-\lambda_1^{k} & \lambda_2^{k-1}-\lambda_1^{k-1}+\lambda_2^{k}-\lambda_1^{k}\\
\lambda_2^{k+1}-\lambda_1^{k+1} & \lambda_2^{k}-\lambda_1^{k}+\lambda_2^{k+1}-\lambda_1^{k+1}
\end{array}\right)
\end{align*}
So this is Fibonacci I guess? And therefore equivalent to:
\begin{align*}
\frac{1}{\sqrt{5}}\left(\begin{array}{cc}
\lambda_2^{k}-\lambda_1^{k}& \lambda_2^{k+1}-\lambda_1^{k+1}\\
\lambda_2^{k+1}-\lambda_1^{k+1} & \lambda_2^{k+2}-\lambda_1^{k+2}
\end{array}\right)
\end{align*}
(Revision 2)
$A^k=P\Lambda^{k}P^{-1}$:
\begin{align*}
\left(\begin{array}{cc}
-\lambda_2& -\lambda_1\\
1 & 1
\end{array}\right)\cdot\left(\begin{array}{cc}
\lambda_1^k& 0\\
0 & \lambda_2^k
\end{array}\right)\cdot\frac{1}{\sqrt{5}}\left(\begin{array}{cc}
-1& -\lambda_1\\
1 & \lambda_2
\end{array}\right)=\frac{1}{\sqrt{5}}\left(\begin{array}{cc}
\lambda_1^k\lambda_2-\lambda_1\lambda_2^k& \lambda_1^{k+1}\lambda_2-\lambda_1\lambda_2^{k+1}\\
\lambda_2^k-\lambda_1^k & \lambda_2^{k+1}-\lambda_1^{k+1}
\end{array}\right)
\end{align*}
Is this the right way?

1) A straightforward proof which is more natural than recursion, in my opinion (for a recursion proof see 2).)
Use diagonalization identity $A=P\Lambda P^{-1}$ from which $A^k=P\Lambda^kP^{-1} \ \ (1)$
where $\Lambda$ is the diagonal matrix diag$(\lambda_1,\lambda_2)$.
Here is an extension of my first explanation:
Indeed the columns of matrix $P$ are eigenvectors associated with $\lambda_1$ ans $\lambda_2$ in this order ; we can take 
$P=\left(\begin{array}{cc}
-\lambda_2 & -\lambda_1\\
1 & 1
\end{array}\right)$ with 
$P^{-1}=\dfrac{1}{\sqrt{5}}\left(\begin{array}{cc}
-1 &-\lambda_1 \\
1 &\lambda_2
\end{array}\right)$.
Plugging these expressions into formula (1) gives the answer.
2) @dk20, as you asked for a recursion proof, I add it to the previous text:
One wants to prove that, for any $k>1$, matrices
$A=\left(\begin{array}{cc}
0 & 1 \\
1 & 1
\end{array}\right)\left(\begin{array}{cc}
\lambda_2^{k-1}-\lambda_1^{k-1} & \lambda_2^{k}-\lambda_1^{k}\\
\lambda_2^{k}-\lambda_1^{k} & \lambda_2^{k+1}-\lambda_1^{k+1}
\end{array}\right)$ 
and $B=\left(\begin{array}{cc}
\lambda_2^{k}-\lambda_1^{k} & \lambda_2^{k+1}-\lambda_1^{k+1}\\
\lambda_2^{k+1}-\lambda_1^{k+1} & \lambda_2^{k+2}-\lambda_1^{k+2}
\end{array}\right)$
are identical.
It is clear that $A_{1j}=B_{1j}$ ($j=1,2$:coefficients of the first line).
Let us now prove that $A_{21}=B_{21}$ (bottom left coefficients), i.e.,
$\lambda_2^{k-1}-\lambda_1^{k-1} + \lambda_2^{k}-\lambda_1^{k}=\lambda_2^{k+1}-\lambda_1^{k+1}$
This equation is equivalent to the following one:
$\lambda_1^{k-1}(1+\lambda_1-\lambda_1^2)=\lambda_2^{k-1}(1+\lambda_2-\lambda_2^2)$
which is evidently fulfilled because it boils down to $0=0$ ; indeed, $\lambda_1$ and $\lambda_2$ are both roots of the quadratic equation $x^2-x-1=0$ ($\lambda_2$ is the "golden number").
The reason why $A_{22}=B_{22}$ is identical (change $k$ into $k+1$).
