We will prove more general claim: for permutations $\sigma\in S(X)$ of a set $X=\{x_1, x_2, \ldots, x_n\}$ the sum is equal to $\frac{1}{x_1x_2\ldots x_n}$. To prove that we will induct on $n$. The base case $n=1$ is trivial. For the induction step consider all permutations of $Y=X\cup\{x_{n+1}\}$ and divide them into $n+1$ groups in th following way: the $i$-th group will contain permutations $\sigma\in S(Y)$ which satisfy $\sigma(x_{n+1})=x_i$. Now, applying the inductive assumption gives that $$ \sum_{\sigma\in S(Y)}\frac{1}{(a_1)(a_1+a_2)(a_1+a_2+a_3)\dots(a_1+a_2+\dots+a_n)}=\sum_{i=1}^{n+1}\frac{1}{\prod\limits_{j\neq i}x_j}\cdot\frac{1}{x_1+x_2+\ldots+x_{n+1}}=\frac{1}{\prod\limits_{i=1}^{n+1}x_i}, $$as desired.