Given that $$f(k) = \prod_{i=1}^k a_i + \sum_{b=1}^{k-1} (1-a_{k-b}) \prod_{i=1}^b a_{k-b+i}$$ for all $k$ where $(a_1, a_2, a_3, \ldots )$ are random constants, prove that: $$f(k+1) = f(k) \cdot a_{k+1} + (1 - a_k)a_{k+1}$$
So far, my current work is:
$$f(k+1) = \prod_{i=1}^{k+1} a_i + \sum_{b=1}^{k} (1-a_{k-b+1}) \prod_{i=1}^b a_{k-b+i+1} = a_{k+1}\prod_{i=1}^{k+1} a_i + \sum_{b=1}^{k} (1-a_{k-b+1}) \prod_{i=1}^b a_{k-b+i+1}.$$
So, we have to show that: $$a_{k+1}\prod_{i=1}^{k+1} a_i + \sum_{b=1}^{k} (1-a_{k-b+1}) \prod_{i=1}^b a_{k-b+i+1} = a_{k+1}\prod_{i=1}^{k+1} a_i + a_{k+1}\sum_{b=1}^{k-1} (1-a_{k-b}) \prod_{i=1}^b a_{k-b+i} + (1-a_k)(a_{k+1}).$$
The $a_{k+1}\prod_{i=1}^{k+1} a_i$ cancel out on both sides, leaving us to prove: $$\sum_{b=1}^{k} (1-a_{k-b+1}) \prod_{i=1}^b a_{k-b+i+1} = a_{k+1}\sum_{b=1}^{k-1} (1-a_{k-b}) \prod_{i=1}^b a_{k-b+i} + (1-a_k)(a_{k+1}).$$
Turning the max value of $i$ on the first summation to $k-1$ and changing the second series of products a bit to match the first one gives:
\begin{align} \sum_{b=1}^{k-1} (1-a_{k-b+1}) \prod_{i=1}^b a_{k-b+i+1} + (1-a_1)\prod_{i=1}^k a_{k-b+i+1} = &\\ a_{k+1} \sum_{b=1}^{k-1} \left( (1-a_{k-b}) \prod_{i=1}^b a_{k-b+i+1} \cdot \frac{a_{k-b+1}}{a_{k+1}}\right) + (1-a_k)(a_{k+1}). \end{align}
We can cancel out the $a_{k+1}$ terms on the RHS, leaving us to prove: $$\sum_{b=1}^{k-1} (1-a_{k-b+1}) \prod_{i=1}^b a_{k-b+i+1} + (1-a_1)\prod_{i=1}^k a_{k-b+i+1} = \sum_{b=1}^{k-1} \left( (1-a_{k-b}) a_{k-b+1} \prod_{i=1}^b a_{k-b+i+1}\right) + (1-a_k)(a_{k+1}).$$
However, I don't know how to continue on from here. Any help would be greatly appreciated.