Proving the binomial series for all real (complex) n using Taylor series

Question

We know the binomial theorem $$(a+b)^n =\sum_{k=1}^{n}{n\choose k}a^{n-k}b^k$$ can be proved using mathematical induction $\forall n\in N$. What I've been trying is to prove it for all real $n$ (or even complex $n$) as follows: Consider the function $$f(b)=(a+b)^n$$ then the $k$-th derivative is given by: $$f^{(k)}(b)=n(n-1)(n-2)\cdots(n-k+1)(a+b)^{n-k}$$ Considering now the taylor series of the function $f$ at $0$ one gets: $$f(b)=(a+b)^n=\sum_{k=0}^\infty\frac{f^{(k)}(0)}{k!}(b-0)^k $$ but $f^{(k)}(0)=n(n-1)(n-2)\cdots(n-k+1)a^{n-k}$ and complementing this with $\frac{1}{k!}$ gets the binomial coefficient $n\choose k$ therefore $$(a+b)^n=\sum_{k=0}^\infty {n\choose k}a^{n-k}b^k$$ Which would be valid since for $n\in N$ the terms with $k>n$ are equal to zero so this would terminate at $n$ (and is the original binomial theorem). For $n\notin N$ this would turn into infinite series. Is this a good approach and would this also prove that the binomial theorem holds for all $n\in C$?