Power Rule for Derivatives/Natural Number Index/Proof by Induction

Proof
We will use the notation $D \map f x = \map {f'} x$ as it is convenient.

Let $n = 0$.

Then:
 * $\forall x \in \R: x^n = 1$

Thus $\map f x$ is the constant function $\map {f_1} x$ on $\R$.

Thus from Derivative of Constant, $D \map f x = \map D {x^0} = 0 x^{-1}$, except where $x = 0$.

So the result holds for $n = 0$.

Let $n = 1$.

Then:
 * $\forall x \in \R: \map f x = x^n = x$

Then from Derivative of Identity Function:
 * $\map D x = 1 = 1 \cdot x^{1 - 1}$

So the result holds for $n = 1$.

Now assume $\map D {x^k} = k x^{k - 1}$.

Then by the Product Rule for Derivatives:
 * $\map D {x^{k + 1} } = \map D {x^k x} = x^k \map D x + \map D {x^k} x = x^k \cdot 1 + k x^{k - 1} x = \paren {k + 1} x^k$

The result follows by induction.