I've been reading some papers recently on higher-order Fibonacci Numbers (aka "n-bonacci"),
and I found to be common practice to define them by the recurrence
$$
F_{\,n} ^{\left( m \right)} = \sum\limits_{k = 1}^m {F_{\,n - k} ^{\left( m \right)} }
$$
so that there is common agreement on that $m=2$ be the standard Fibonacci,
and assigning as initial values the $m$-tuple
$$
\left( {\underbrace {0,0, \cdots ,0}_{m - 1\;{\rm zeros}},1} \right)
$$
The only difference I found is that the $m$-tuple
- by some sources (e.g 1) is made to start at the index $0$
$$
F_{\,0} ^{\left( m \right)} = F_{\,1} ^{\left( m \right)} = \cdots = F_{\,m - 2} ^{\left( m \right)} = 0,\quad F_{\,m - 1} ^{\left( m \right)} = 1
$$
- others (e.g. 2) prefer to fix the one at $n=1$ and so put
$$
F_{\, - m + 2} ^{\left( m \right)} = F_{ - m + 3} ^{\left( m \right)} = \cdots = F_{\,0} ^{\left( m \right)} = 0,\quad F_{\,1} ^{\left( m \right)} = 1
$$
The difference is just a shift in the lower index.
I personally prefer the second setting since it provides a simpler extension of the o.g.f.
$$
\eqalign{
& G\left( {z,2} \right) = {z \over {1 - z - z^{\,2} }} = {z \over {1 - z\left( {1 + z} \right)}} = {z \over {1 - z{{1 - z^{\,2} } \over {1 - z}}}}\quad \Rightarrow \cr
& \Rightarrow \quad G\left( {z,m} \right) = {z \over {1 - z\left( {1 + z + \cdots + z^{\,m - 1} } \right)}} = {z \over {1 - z{{1 - z^{\,m} } \over {1 - z}}}} \cr}
$$
Best Answer
One method uses matrices. Let $\,A = (^{0\, 1}_{1\, 1}).\,$ A simple induction proof shows that $$ A^n = \pmatrix{ F_{n-1} \ F_n \\ F_n \ F_{n+1} }. $$ The equation $\,A^{m+n} = A^m\ A^n\,$ implies your equation for $\,F_{m+n}.\,$ Use the equation $\,A^{m+n+k} = A^m\ A^n\ A^k\,$ for a similar Fibonacci equation and so on.