As a simpler example, consider the iterated sum with only 2 indices,

(The case with just one index is pretty simple, as it reduces to ζ(2) = π²/6.)
Let

Differentiating and multiplying by x, we get


By the fundamental theorem of calculus (observing that letting x = 0 in the sum makes it vanish), we have

If we let x approach 1 from below, f(x) will converge to the double sum and

In the integral, substitute
, use the power series expansion for 1/(1 - x), and integrate by parts twice.

We can generalize this method to k indices to show that

Then the sum we want is
