Proving that $(ax^n)' = nax^{n-1}$ using the definition of the derivative Given $f(x) = ax^n$, we have that 
$$f'(x) =  \lim_{h\to 0}  \frac{a(x+h)^n-ax^n}{h}$$
While its easy to prove this by induction by already implying that we know that $(ax^n)' = nax^{n-1}$, is there a simple way to solve the limit given above?
 A: The easiest way to do this is by simply multiplying out the $(x+h)^n$:
$$
(x+h)^n=\sum_{k=0}^{n}\binom{n}{k}x^kh^{n-k},
$$
where $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ is the binomial coefficient. Then 
$$
\frac{a(x+h)^n-ax^n}{h}=\frac{a}{h}\sum_{k=0}^{n-1}\binom{n}{k}x^kh^{n-k}=a\sum_{k=0}^{n-1}\binom{n}{k}x^kh^{n-k-1}.
$$
Noting that the $k=n-1$ term does not involve an $h$, while all other terms involve a positive power of $h$, you can compute the desired limit.
A: If you want to prove the power rule for any value of n ,rational or irrational, I am afraid you need to resort to e-powers. Once you learned about transcedental functions, it becomes very easy to solve. 
A: Hint:
$$(x+h)^n=\sum_{m=0}^nc(n,m)x^mh^{n-m}$$
where $$c(n,m)=\frac{n!}{m!(n-m)!}$$
A: Using the Binomial Formula is the standard way to do this,
but another approach is to let $t=x+h$ to get
$\displaystyle f^{\prime}(x)=\lim_{h\to0}\frac{a(x+h)^n-ax^n}{h}=\lim_{t\to x}\frac{a(t^n-x^n)}{t-x}=\lim_{t\to x}\frac{a(t-x)(t^{n-1}+t^{n-2}x+t^{n-3}x^2+\cdots+tx^{n-2}+x^{n-1})}{t-x}=$
$\displaystyle\lim_{t\to x}a(t^{n-1}+t^{n-2}x+t^{n-3}x^2+\cdots+tx^{n-2}+x^{n-1})=anx^{n-1}.$
