Mathematical Induction: Power Rule for Differentiation

[BrianMath]

Homework Statement

Prove that
\frac{d}{dz}z^n = nz^{n-1}\;\;\; \forall n\in\mathbb{N}
using the Product Rule for differentiation and mathematical induction.

Homework Equations

\frac{d}{dz} f(z) = \lim_{\Delta z\to 0} \frac{f(z+\Delta z) - f(z)}{\Delta z}
\frac{d}{dz}[f(z)g(z)] = f\,'(z)g(z) + g'(z)f(z)

The Attempt at a Solution

Let n = 1:
\frac{d}{dz} z = \lim_{\Delta z \to 0} \frac{z + \Delta z - z}{\Delta z} = \lim_{\Delta z \to 0} \frac{\Delta z}{\Delta z} = \lim_{\Delta z \to 0} 1 = 1

Assume true for n = k:
\frac{d}{dz} z^k = kz^{k-1}

Let n = k + 1:
\frac{d}{dz} z^{k+1} = \frac{d}{dz}[z^k\cdot z] = \frac{d}{dz} (z^k) \cdot z + \frac{d}{dz} (z) \cdot z^k = kz^{k-1}\cdot z + z^k = kz^k + z^k = (k+1)z^k = (k+1)z^{(k+1)-1}

Like my previous topic, I'm pretty sure I have the proof correct, but I need to make sure that it is written out correctly. I don't want to be teaching myself the wrong way to lay out proofs.

 

[LCKurtz]

Looks fine to me.