Is it true that: $\sum _{i=0}^\infty\left( \prod _{j=0}^{i-1}p_{j}-\prod _{j=0}^{i}p_{j} \right)=\left(1-\prod _{j=0}^{\infty }p_{j}\right)$? Okay, I'm trying to prove some exercise, and after hours and hours of trying, I think I can show it, if the following is true:
$$\sum _{i=0}^\infty\left( \prod _{j=0}^{i-1}p_{j}-\prod _{j=0}^{i}p_{j} \right)=\left(1-\prod _{j=0}^{\infty }p_{j}\right)$$
or equivalently
$$\sum _{i=0}^\infty (1-p_{i})\prod _{j=0}^{i-1}p_{j}=\left(1-\prod _{j=0}^{\infty }p_{j}\right)$$
or equivalently
$$\sum _{i=0}^\infty (1-p_{i})p_{i-1}\ldots p_0=\left(1-\prod _{j=0}^{\infty }p_{j}\right)$$
Where $0<p_i<1$ and also $0<\prod _{j=0}^{\infty}p_{j}<1$. So $p_i$ is converging sufficiently rapidly to $1$ to make sure that $\prod _{j=0}^{\infty}p_{j}>0$.
So I really hope that this is the case, but I'm not able to show it. I would appreciate it if someone here can show to me, why or why not this is true.

Hint: Prove by induction on $K$ that
$$
\sum_{i=0}^K \bigg( \prod_{j=0}^{i-1} p_j - \prod_{j=0}^{i} p_j \bigg) = 1 - \prod_{j=0}^K p_j.
$$
(Certainly some condition on the $p_j$ is needed for the product $\prod_{j=0}^\infty p_j$ to converge; $0<p_j<1$ isn't enough.)
