Statement: For any integer $\displaystyle n\geq 0, (x^n)'=0 $
Proof
Base Case: For $\displaystyle n=0, x^0=1, so \, (x^0)'=(1)'=0 $
Induction hypothesis: Assume that for some $\displaystyle k, (x^i)'=0 $ for all $\displaystyle 0 \leq i \leq k$
Induction step: Consider $\displaystyle x^{k+1} $. Note that $\displaystyle x^{k+1} = x* x^k$
Using product rule, $\displaystyle (x^{k+1})'=(x*x^k)'=x*(x^k)' + (x)'*x^k $
By induction hypothesis, $\displaystyle ( x^k)'=0 $ and $\displaystyle (x)'=0 $. Therefore, $\displaystyle (x^{k+1})'=0$. So the result holds by induction
I know this statement is obviously false, but I can't seem to find anything wrong with the proof that is provided.
I looked through my notes and it looks like every step of the induction process is followed correctly.
Can someone please help me figure out which part is wrong? or give me a hint?