summation - Why $sum_{i=0}^infty i^2p^{i-1} = sum_{i=0}^infty i(frac{d}{dp}p^{i-1})$?

I'm going through a proof right now and am having trouble figuring out the math behind one line. It says:

$$\sum_{i=0}^\infty i^2p^{i-1} = \sum_{i=0}^\infty i(\frac{d}{dp}p^i)$$

I know this question is vague but can anybody explain why this is the case?

Answer

First of all, the summation notation is unnecessary and irrelevant. Remove it and we get
$$i^2p^{i-1}=i\big(\frac{d}{dp}p^i\big)$$
Okay. Recall the formula
$$\frac{d}{dx}x^k=kx^{k-1}, k\ne0$$
Then we have that

$$\frac{d}{dp}p^{i}=ip^{i-1}, i\ne0$$
and so, by substitution, we have
$$i\big(\frac{d}{dp}p^i\big)=i\big(\frac{d}{dp}p^i\big)$$
$$i\big(\frac{d}{dp}p^i\big)=i(ip^{i-1})$$
$$i\big(\frac{d}{dp}p^i\big)=i^2p^{i-1}$$
Does that answer your question?