Proving that, if $f(k)=\prod_{i=1}^ka_i+\sum_{b=1}^{k-1}(1-a_{k-b})\prod_{i=1}^ba_{k-b+i}$, then $f(k+1)=f(k)\cdot a_{k+1}+(1 - a_k)a_{k+1}$ 
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.
 A: We can considerably simplify $f(k)$ which makes the proof easy.

We obtain
\begin{align*}
\color{blue}{f(k)}&=\prod_{i=1}^k a_i+\sum_{b=1}^{k-1}\left(1-a_{k-b}\right)\prod_{i=1}^b a_{k-b+i}\\
&=\prod_{i=1}^k a_i+\sum_{b=1}^{k-1}\left(1-a_{b}\right)\prod_{i=1}^{k-b} a_{b+i}\tag{1}\\
&=\prod_{i=1}^k a_i+\sum_{b=1}^{k-1}\prod_{i=1}^{k-b} a_{b+i}-\sum_{b=1}^{k-1}a_b\prod_{i=1}^{k-b} a_{b+i}\\
&=\prod_{i=1}^k a_i+\sum_{b=2}^{k}\prod_{i=1}^{k-b+1} a_{b-1+i}-\sum_{b=1}^{k-1}\prod_{i=0}^{k-b} a_{b+i}\tag{2}\\
&=\prod_{i=1}^k a_i+\sum_{b=2}^{k}\prod_{i=0}^{k-b} a_{b+i}-\sum_{b=1}^{k-1}\prod_{i=0}^{k-b} a_{b+i}\tag{3}\\
&=\prod_{i=1}^k a_i+a_k-\prod_{i=0}^{k-1} a_{i+1}\tag{4}\\
&=\prod_{i=1}^k a_i+a_k-\prod_{i=1}^{k} a_{i}\tag{5}\\
&\,\,\color{blue}{=a_k}\tag{6}
\end{align*}

Comment:

*

*In (1) we change the order of summation of the sum $b\to k-b$.


*In (2) we shift the index of $b$ by one in the left sum and merge the factor $a_b$ into the product of the right sum.


*In (3) we shift the index of the product in the left sum by one and observe the sums are telescoping.


*In (4) we do the telescoping.


*In (5) we shift the index of the right-hand product and simplify in the next line.

With $f(k)=a_k$ the proof is simple, since we have
\begin{align*}
\color{blue}{f(k) a_{k+1} + (1 - a_k)a_{k+1}}&= a_ka_{k+1}+(1-a_k)a_{k+1}=a_{k+1}\\
&\color{blue}{=f(k+1)}
\end{align*}
and the claim follows.

