Prove $\frac{d}{dx} x^n=nx^{n-1} : \forall n\in \mathbb{Z}_{+}$ by induction Problem
Prove $$\frac{d}{dx} x^n=nx^{n-1} : \forall n\in \mathbb{Z}_{+}$$ by mathematical induction.
 Attempt to solve 
Base case
when $n=1$
$$ \frac{d}{dx} x^1 = 1 \cdot x ^{0}=1 $$
which is true
Induction step
$$\frac{d}{dx}x^{n+1}=(n+1)x^{n+1-1}= (n+1)x^{n}$$
At this point not quite sure how to prove this with induction without proving operator $\frac{d}{dx}$ with 
$$ \frac{d}{dx}f(x)=\lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h} $$
and then proving existence of such limit with:
$$ 0<|x-a|< \delta \implies|\frac{f(a+h)-f(a)}{h}-\frac{d}{dx}f(a)| < \epsilon
$$
and then we can arrive at implication that $$\frac{d}{dx}x^n=nx^{n-1} \implies \frac{d}{dx}x^{n+1}=(n+1)x^n$$
Most likely there is easier by induction which is capable of showing that $$ \frac{d}{dx}x^n=nx^{n-1} $$ is applicable $\forall n \in \mathbb{Z}_+$?

Assume that we have 
$(*) \quad \frac{d}{dx} x^n=nx^{n-1}$ 
for some $ n\in \mathbb{Z}_{+}$.
For the induction step use $(*)$ and the product rule !
