Here's a question:
Derive $$\frac{\mathrm{d}}{\mathrm{d}z} z^n = nz^{n-1},$$ when n is a positive integer by using mathematical induction and and the derivative of a product of two functions $$\frac{\mathrm{d}}{\mathrm{d}z} f(z)g(z) = f(z)g'(z) + f'(z)g(z).$$
Here's what I did. Is this mathematical induction performed correctly?
Base case: n = 1 $$\frac{\mathrm{d}}{\mathrm{d}z} z^1 = 1z^{1-1}$$ $$1 = 1$$
Base case is true.
Inductive step: Assume true for $n = k$, show true for $k + 1$.
So show: $$\frac{\mathrm{d}}{\mathrm{d}z} z^{(k+1)} = (k+1)z^{(k+1) - 1} = (k+1)z^k$$
$$\frac{\mathrm{d}}{\mathrm{d}z} z^{(k+1)} = \frac{\mathrm{d}}{\mathrm{d}z} z^kz.$$
Using derivative of a product of two functions: $$\frac{\mathrm{d}}{\mathrm{d}z} z^kz =z^k\cdot 1 + kz^{k-1}\cdot z$$ $$ =z^k + kz^k$$ $$ = (k+1)z^k$$
So as the base case holds and the inductive step holds, this means the original statement is valid?