calculus - Why $sum_{n=0}^{infty}(n+1)5^nx^n=frac{1}{(1-5x)^2}$

Why $$\sum_{n=0}^{\infty}(n+1)5^nx^n=\frac{1}{(1-5x)^2}?$$

I know that $\sum_{n=0}^{\infty}x^n=\dfrac{1}{1-x}$, so by the same token, $\sum_{n=0}^{\infty}5^nx^n=\dfrac{1}{1-5x}$.

Thus
$$\left(\frac{1}{1-5x}\right)^2=\frac{1}{(1-5x)^2} = \left(\sum_{n=0}^{\infty}5^nx^n\right)^2.$$

But why is $\big(\sum_{n=0}^{\infty}5^nx^n\big)^2=\sum_{n=0}^{\infty}(n+1)5^nx^n$?

Assuming $x$ is small enough so that the sum converges.

Answer

Note that
$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n,$$
and thus
$$\frac{1}{(1-x)^2}=\left(\frac{1}{1-x}\right)'=\sum_{n=1}^\infty nx^{n-1}=\sum_{n=0}^\infty (n+1)x^{n},$$
and hence

$$\frac{1}{(1-5x)^2}=\sum_{n=0}^\infty (n+1)(5x)^{n}.$$