Fibonacci addition law $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$

Question: Let $F_n$ the sequence of Fibonacci numbers, given by $F_0 = 0, F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for $n \geq 2$. Show for $n, m \in \mathbb{N}$: $$F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$$

My (very limited) attempt so far: after creating a small list of the values $F_0=0, F_1=1, F_2=1, F_3=2, F_4=3, F_5=5, F_6=8, F_7=13, F_8=21, F_9=34, F_{10}=55$ i can see that yes it does seem to work for instance $F_{6+3}=F_5 F_3 +F_6 F_4 = 10 +24 = 34 = F_9$. However, I really don't know where to begin as showing that this must hold in general terms. Should I be looking to use limits? Or perhaps induction? What is the best way to solve this?

With Fibonacci numbers usually there are multiple ways of proving identities.

One way (which is one of my favourites) to prove your identity is the following:

Consider the following problem:

A person climbs up $\displaystyle n$ steps, by taking either one step, or two steps at a time.

The total number of ways the person can climb up all the $\displaystyle n$ steps is $\displaystyle F_{n+1}$ (Why?)

Now consider climbing $\displaystyle m+n-1$ steps and split into the cases when the person lands on step $\displaystyle n$ and the cases when the person lands on step $\displaystyle n-1$ and takes two steps at that point (and so does not land on step $\displaystyle n$ in those cases). These two cases cover all possibilities, and so we have:

$$\displaystyle F_{m+n} = F_{n+1}F_{m} + F_{n}F_{m-1}$$

If pressed, I'd nuke with Binet's formula myself, but here's another approach. By an easy induction, if $$A=\pmatrix{0&1\\\\ 1&1}$$ then $$A^n=\pmatrix{F_{n-1}&F_n\\\\ F_n&F_{n+1}}.$$ Comparing the top right entry in the equation $A^m A^n=A^{m+n}$ gives $$F_{m-1}F_n+F_m F_{n+1}=F_{m+n}.$$

Hint $\ $ If we put the Fibonacci recurrence into matrix form then the result is obvious, viz.

$$ M^n\ :=\ \left[\begin{array}{rr} \!\!1 & 1 \\\ \!\!1 & 0 \end{array}\right]^{\large n} =\, \left[\begin{array}{cc} F_{n+1} &\!\! F_n \\\ F_n &\!\! F_{n-1} \end{array}\right] $$

Now comparing the entries of $\ M^{n+m} = M^n \ M^m\,$ immediately yields the Fibonacci addition law.

Remark $\ M$ is the shift operator, i.e. $\,(F_{n−1},F_n)M^t = (F_n,F_{n+1}).\,$ The same idea works for any linear recurrence.

Fix $m \in \mathbb{N}$. We shall use induction on $n$. For $n=1$, the RHS of the equation becomes $$F_{m-1}F_1 + F_mF_2 = F_{m-1} + F_{m}$$, which is equal to $F_{m+1}$. When $n=2$, the equation is also true.( I hope you can prove this!).

Now assume, that the result is true for $k=3,4, \cdots , n$. We want to show that the result is true for $k=n+1$. $$ \text{For} \ k=n-1 \ \text{we have} \quad F_{m+n-1} = F_{m-1}F_{n-1} + F_mF_n$$ and $$ \text{For} \ k = n \ \text{we have} \quad F_{m+n}=F_{m-1}F_{n} + F_{m}F_{n+1}$$ Adding both the sides you will get $$F_{m+n-1} + F_{m+n} = F_{m+n-1} = F_{m-1}F_{n+1} + F_{m}F_{n+2}$$

Oh, i reversed the notations. This is for proving $$F_{m+n} = F_{m-1}F_{n} + F_{m}F_{n+1}$$