Power function derivative

A power function is a function of the form $f(x)=x^a$ . The derivative is defined as \begin{align} f'(x) &= \lim_{h\rightarrow 0}\left(\frac{f(x+h)-f(x)}{h} \right)\\ &= \lim_{h\rightarrow 0}\left(\frac{\left(x+h\right)^a - x^a}{h}\right)\\ &= x^{a}\cdot \lim_{h\rightarrow 0}\left(\frac{\left(1+\frac{h}{x}\right)^a - 1}{h}\right)\,. \end{align} For $\left|\frac{h}{x}\right|<1$ the term $(1+\frac{h}{x})^a$ can be expanded using the generalized binomial coefficient as \begin{align} \left(1+\frac{h}{x}\right)^a &= \sum_{k=0}^{\infty} \left(\begin{array}{c}a \\ k\end{array}\right) \left(\frac{h}{x}\right)^k\,. \end{align} where $a\in\mathbb C$, and the generalized binomial coefficient is \begin{align} \left(\begin{array}{c}a \\ k\end{array}\right) &= \frac{a^{\underline{k}}}{k!} = \frac{1}{k!}\prod_{i=0}^{k-1}(a-i)\,. \end{align} The limit becomes \begin{align} f'(x) &= x^{a}\cdot \lim_{h\rightarrow 0}\left(\frac{\sum\limits_{k=0}^{\infty} \left(\frac{1}{k!}\left(\prod\limits_{i=0}^{k-1}(a-i)\right) \left(\frac{h}{x}\right)^k\right)- 1}{h}\right) \end{align} Since the term $k=0$ in the sum makes the iterated product empty this term gives \begin{align} \left(\frac{1}{0!}\left(\prod\limits_{\emptyset}(a-i)\right) \left(\frac{h}{x}\right)^0\right) = 1 \end{align} and the limit is slightly simplified to \begin{align} f'(x) &= x^{a}\cdot \lim_{h\rightarrow 0}\left(\frac{\sum\limits_{k=1}^{\infty} \left(\frac{1}{k!}\left(\prod\limits_{i=0}^{k-1}(a-i)\right) \left(\frac{h}{x}\right)^k\right)}{h}\right)\\ &= x^{a}\cdot \lim_{h\rightarrow 0}\left(\sum\limits_{k=1}^{\infty} \left(\frac{1}{k!}\cdot \left(\prod\limits_{i=0}^{k-1}(a-i)\right)\cdot \frac{h^{k-1}}{x^k}\right)\right) \end{align} In the limit $h\rightarrow 0$ the factor $h^{k-1}$ approaches zero for $k>1$, while it approaches $1$ for $k=1$: \begin{align} f'(x) &= x^{a}\cdot \sum\limits_{k=1}^{\infty} \left(\frac{1}{k!}\cdot \left(\prod\limits_{i=0}^{k-1}(a-i)\right)\cdot \frac{\delta_{k,1}}{x^k}\right)\\ &= x^{a}\cdot \frac{1}{1!}\cdot \left(a-0\right)\cdot \frac{1}{x^1}\\ &= a\cdot x^{a-1}\,. \end{align} We've thus recovered the well known derivative $f'(x)=a\cdot x^{a-1}$ for the power function $f(x)=x^a$.