Series identity

[Easty]

1. The problem statement, all variables and given/known data[/b]

f _{a} (z) is defined as

f(z) = 1 + az + \frac{a(a-1)}{2!}z^{2}+....+\frac{a(a-1)(a-2)...(a-n+1)}{n!}z^{n} + ......

where a is constant

Show that for any a,b

f _{a+b} (z)= f _{a}(z)f _{b}(z)

2. Relevant equations


3. The attempt at a solution

I've tried starting directly from f_a+f_b and trying to show it is equivalent to f_ab and vice versa but i keep getting stuck with the last general term, im thinking there is a better way to approach this question but i cant see it.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

[CompuChip]

Is n a fixed number, or is it an infinite series?
Perhaps it helps if you write the numerators in terms of factorials as well, perhaps you will even recognize some binomial coefficients ;)

[HallsofIvy]

Or look at the derivatives: fa(0)= 1, fa'(0)= a, fa"(0)= a(a-1) and, in general
\frac{d^n f^a}{dx^n} (0)= a(a-1)\cdot\cdot\cdot (a-n+1)
and of course,
\frac{d^n f^b}{dx^n} (0)= b(b-1)\cdot\cdot\cdot (b-n+1)

Try using the product rule, extended to higher derivatives:
\frac{d^n fg}{d x^n}= \sum_{i=0}^n \left(\begin{array}{c}n \\ i\end{array}\right)\frac{d^{n-i}f}{dx}\frac{d^ig}{dx}

[Easty]

Thank you Hallsofivy. Once i took your advice the answer was quite simple to obtain, it was a nice way to approach the problem that i would have never seen.