For math enthusiasts we present two proofs.

In the first proof we assume that $p$ is a positive integer and we use the following rules for differences of powers:

$$a^p-b^p=(a-b)(a^{p-1}+a^{p-2}b+a^{p-3}b^2+\cdots+ab^{p-2}+b^{p-1})$$ Then the derivative $f'(a)$ at the point$x=a$ for some $a$ satisfies:
$$\begin{aligned}f'(a) &=\lim_{h\to 0}\frac{(a+h)^p-a^p}{h}\\[0.25cm] &=\lim_{h\to 0}\frac{h\bigl((a+h)^{p-1}+(a+h)^{p-2}a+(a+h)^{p-3}a^2+\cdots+(a+h)a^{p-2}+a^{p-1}\bigr)}{h}\\[0.25cm] &=\lim_{h\to 0} \bigl((a+h)^{p-1}+(a+h)^{p-2}a+(a+h)^{p-3}a^2+\cdots+(a+h)a^{p-2}+a^{p-1}\bigr)\\[0.25cm] &=a^{p-1}+a^{p-2}a+a^{p-3}a^2+\cdots+a\cdot a^{p-2}+a^{p-1}\quad (p\text{ terms})\\[0.25cm] &=p\cdot a^{p-1}\end{aligned}$$

In the second proof we anticipate on the chain rule for differentiation and the derivative of the natural logarithm, but the advantage is that that now $p$ does not have to be a positive integer. If $$f(x)=x^p$$ then $$\begin{aligned}\ln\bigl(f(x)\bigr)&= \ln(x^p)\\[0.25cm] &= p\cdot \ln(x)\end{aligned}$$ From the chain rule for differentiation and the derivative of the natural logarithm it follows that $$\frac{f'(x)}{f(x)}=p\cdot \frac{1}{x}$$ So: $$\begin{aligned}f'(x)&=p\cdot \frac{f(x)}{x}\\[0.25cm] &= p\cdot \frac{x^p}{x}\\[0.25cm] &=p\cdot x^{p-1}\end{aligned}$$