Let $a_0 = 1$. We prove the following identity:
$$\sum_{k=1}^{n} \prod_{j=1}^{k} \frac{a_{j-1}}{1+a_j} = 1 - \prod_{j=1}^{n} \frac{a_j}{1+a_j} \qquad \textcircled{1}$$
by induction on $n$.

It is evident that $\textcircled{1}$ is true for $n = 1$. Suppose that $\textcircled{1}$ is true for $n-1$, $n \ge 2$, then for $n$,
$$\begin{aligned} \sum_{k=1}^{n} \prod_{j=1}^{k} \frac{a_{j-1}}{1+a_j} &= \sum_{k=1}^{n-1} \prod_{j=1}^{k} \frac{a_{j-1}}{1+a_j} + \prod_{j=1}^{n} \frac{a_{j-1}}{1+a_j} \\ &= 1 - \prod_{j=1}^{n-1} \frac{a_j}{1+a_j} + \prod_{j=1}^{n} \frac{a_{j-1}}{1+a_j} \\ &= 1 - \prod_{j=1}^{n} \frac{a_j}{1+a_j}. \end{aligned} \quad \square$$