Simplifying iterated sums where the upper bound is the next index

Consider first the iterated integral $I=m!int_0^xint_0^{x_1} … int_0^{x_{m-1}} prod_{i=1}^m dx_i f(x_i)$.

By permuting the dummy indices, we can reorder $x_1$ to $x_m$ to find new expressions for $I$. By adding all of these expressions together and dividing by $m!$, we recover:

$$m!int_0^xint_0^{x_1} … int_0^{x_{m-1}}prod_{i=1}^m dx_i f(x_i) = left(int_0^x dy f(y)right)^m.$$

Now consider the sum: $$S = m! sum_{n_1=0}^n sum_{n_2=0}^{n_1} … sum_{n_m=0}^{n_{m-1}} prod_{i=1}^m a_{n_i}$$

Using the same tactic used on the integrals for $S$ for the case of $m=2$, I find that
$$2!sum_{n_1=0}^n sum_{n_2=0}^{n_1} a_{n_1} a_{n_2} = left(sum_{j=0}^n a_jright)^2 + left(sum_{j=0}^n a_j^2right)$$
so it is clear that there are necessarily extra terms relative to the integral case.

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How can I simplify the form of $S$? Since the choice of $a_n = 1$ simplifies nicely for general $m$ because of the hockey-stick identity, perhaps there is a straightforward combinatorical path.