Elegant(short and simple) way to Proof $$ \frac{d}{dx}(x^n)$$

Formula = $$ \frac{d}{dx} n^{x} = nx^{n-1} $$
I used to prove this formula using Binomial Theorem. It is pretty tedious if you are weak in algebra.
$$ Proof $$ using Binomial Theorem

Let's plug $$ f(x) = x^{n} $$
$$ f(x) = \lim_{h\to\infty}\frac{(x+h)^{n} - x^{n}}{h} $$
$$ = \lim_{h\to\infty} \frac{\bigg( (x+h)^{n} = \sum_{p=0}^{\infty}\dbinom{n}{p} x^{n-p} h^{p}\bigg) - x^{n}} {h} $$
If you expand and compute the above... you will reach...
$$ = \lim_{h\to\infty} nx^{n-1} + \frac{n(n-1)}{2!}x^{n-2}h+ ........ + nxh^{n-2} + h^{n-1} $$
$$ = nx^{n-1} $$ This completes the proof.

Another elegant way to prove this derivative is using $$ Logarithmic $$ technique. Let's first define $$ y=x^{n} $$

$$ \ln y = \ln x^{n} ...... \underbrace {\ln y = n \ln x }_\textrm{differentiate both sides}$$
$$ \frac{\tilde{y}}{y} = n \frac{1}{x} ..... \tilde{y} = n \frac{1}{x} \otimes y $$ where $$ y = x^{n}$$.

Finally, $$ \frac{d}{dx} n^{x} = nx^{n-1} $$ is proved elegantly.