#### Answer

See proof below.

#### Work Step by Step

$1. \ (\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e})$ Check that the formula holds for n = 1.
$LHS= \displaystyle \frac{d}{dx}[x^{1}]=1\qquad $
(we are given this by the problem text)
$RHS=1\cdot x^{1-1}=1\cdot x^{0}=1\cdot 1=1=LHS$
$\qquad $
yes, it is true when $n=1.$
$2. \ (\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{p})$ Prove that IF the formula holds for any positive integer $n=k$, THEN it also holds for the next integer, $n=k+1$.
Let us assume that for $n=k$,
$ \displaystyle \frac{d}{dx}[x^{k}]=kx^{k-1}$
(we call this the step hypothesis)
For $n=k+1$,
the LHS equals $ \displaystyle \frac{d}{dx}[x^{k+1}]$, which we rewrite as
$LHS= \displaystyle \frac{d}{dx}[x^{k}\cdot x^{1}]\qquad $
Apply the product rule with $\left[\begin{array}{l}
u=x^{k}\\
v=x
\end{array}\right]$
$=(\displaystyle \frac{d}{dx}u)\cdot v+u(\frac{d}{dx}v)$
$=\displaystyle \frac{d}{dx}[x^{k}]\cdot x+x^{k}(\frac{d}{dx}[x])$
The first term contains $\displaystyle \frac{d}{dx}[x^{k}] $ which, by the hypothesis, equals $kx^{k-1}$
In the second term, we are given by the text: $\displaystyle \frac{d}{dx}[x]=1.$
$=kx^{k-1}\cdot x+x^{k}\cdot 1$
$=kx^{k}+x^{k}\qquad$ ... factor out $x^{k}$
$=x^{k}(k+1)$
$=(k+1)\cdot x^{(k+1)-1}=RHS$ for $n=k+1$
which makes the formula true for $n=k+1.$
By mathematical induction, the formula is true for any integer $n.$