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**Jhevon** Let P(n) be the statement we want to prove.

We proceed by induction on $\displaystyle n$, where $\displaystyle n \in \mathbb{N} = \{ 0, 1 , 2, 3, ... \}$

Clearly $\displaystyle P(0)$ is true, since $\displaystyle (x^0)' = 0 = 0x^{0-1}$ ....(of course, $\displaystyle x \ne 0$. if you want x to be anything it wants to be, then take the natural numbers greater than or equal to 1 in your statement)

Assume $\displaystyle P(n)$ is true, we show $\displaystyle P(n + 1)$ is true.

Now, $\displaystyle x^{n + 1} = x \cdot x^n = \cdots $

basically, you want to apply the product rule, and simplify to get $\displaystyle (n + 1)x^n$, which would be $\displaystyle P(n + 1)$