I'm lost on where to start on this proof:

Using the fact that $A^m A^n = A^{m+n}$ , prove the identity

$F\_m F\_n + F\_{m−1} F\_{n−1} = F\_{m+n−1}$

I want to use induction starting with n = 1, but would I also have to make m = 1? I haven't done induction with 2 variables before.

or because of $A^m A^n = A^{m+n}$ should I setup the problem as a matrix (in that case what would the columns/rows be)? 

I tried doing it mathematically however I think my algebra is wrong so I won't post it here. Am I correct to believe that $F\_{m-1} = F\_m\*-1$ is not the same as $2^{n+1} = 2^n\*2$?

Any help would be appreciated, thanks.
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You can actually use induction here. We induct on $n$ proving that the relation holds for all $m$ at each step of the way. For $n=2$, $F\_1 = F\_2 =1$ and the identity $F\_m+F\_{m-1}=F\_{m+1}$ is true for all $m$ by the definition of the Fibonacci sequence. We now have a strong induction hypothesis that the identity holds for values up until $n$, for all $m$. To show that it holds for $n+1$, for all $m$ we note that
$$
F\_m F\_{n+1} + F\_{m-1} F\_n = F\_m(F\_{n-1} + F\_n) + F\_{m-1}(F\_{n-2} + F\_{n-1}) = 
$$
$$
(F\_mF\_n+F\_{m-1}F\_{n-1}) + (F\_mF\_{n-1} + F\_{m-1}F\_{n-2}) = F\_{m+n-1} + F\_{m+n-2} = F\_{m+n}.
$$
This completes the induction.
Fibonacci numbers have a [matrix representation](http://en.wikipedia.org/wiki/Fibonacci_number#Matrix_form):

$$\left( \begin{smallmatrix} F\_{n+1} & F\_n \\ F\_n & F\_{n-1}\end{smallmatrix} \right) = \left( \begin{smallmatrix} 1 & 1 \\ 1 & 0 \end{smallmatrix}\right)^n$$

This is probably what you were meant to use for this problem.