Simplify nested, dependent integral I am trying to simplify an integral that appeared in a problem I'm working on,
$$\int_{s_1=0}^1\int_{s_2=s_1}^1\int_{s_3=s_2}^1\cdots \int_{s_n=s_{n-1}}^1\prod_{i=1}^n f(s_i) ds_n ds_{n-1} \cdots ds_1.$$
Here $n$ is a positive integer and $f(s)$ is a non-negative, non-fractal function without any infinite values, but otherwise arbitrary. I hope that it is possible to simplify the limits, reduce the number of integral signs, or something similar.
I am thinking that it might be possible to expand the product in some basis (Fourier? polynomial?) that would make it possible to explicitly integrate everything afterwards. Does this seem to be a workable strategy? I think it will be quite some work to do any such thing for a general value of $n$, so I thought it would be nice if someone with more math experience could suggest their thoughts on the best course of action.
Edit:
I have worked a bit more on this problem and wanted to add some of my thoughts.
I have tried to express $f(s)$ in the polynomial form (as suggested by Nick Kidman) but I have not managed to make any progress in that direction. Basically, when I try to calculate the integral I quickly need to multiply two such non-trivial polynomials and the expressions for the polynomial coefficient rapidly increase in complexity in some kind of combinatorial explosion.
However, I have found the following interesting things. The integral above can be rewritten as
$$\int_{s_n=0}^1 \int_{s_{n-1}=0}^{s_n} \cdots \int_{s_1=0}^{s_2} \prod_{i=1}^n f(s_i) ds_1 ds_2 \cdots ds_n.$$
It can quite easily be found that this integral evaluates to $n!$ if $f(s)=1$. I have also managed to fully integrate it for $f(s)=Cs^k$, for which I obtain the result
$$\frac{1}{n!} \left( \frac{C}{k+1} \right)^n.$$
Since $\frac{C}{k+1}$ happens to be the mean value for $f(s)$ in the interval from $0$ to $1$, I made the hopeful jump to the following equality that I wanted to test if it holds,
$$\int_{s_n=0}^1 \int_{s_{n-1}=0}^{s_n} \cdots \int_{s_1=0}^{s_2} \prod_{i=1}^n f(s_i) ds_1 ds_2 \cdots ds_n \overset{?}{=} \frac{1}{n!} \left( \int_0^1 f(s) ds \right)^n.$$
I then tried to disprove this equality for $n=2$ and $f(s)=\exp(s)$ and $f(s)=as^2+bs+c$, but both passed. After that I have made some numerical simulations in Matlab for more complicated polynomials and exponentials and sums of both, for $f(s)$, and the expression to the left and the right of the equals sign result in numbers that are suspiciously close to each other (I have not yet cranked up the resolution of my simple code, since that increases runtime a lot).
Any further help is gratefully appreciated. :) For example, if you have any idea how to verify my proposed equality above, it would be nice. Or, if you have a counter-example that disproves it, it would also be great. I will keep working on this problem when time permits.
 A: For a permutation $\pi$ of the $n$ coordinates, let
$$C_\pi = \{ s \in [0,1]^n : 0 \leqslant s_{\pi(1)} \leqslant s_{\pi(2)} \leqslant \dotsc \leqslant s_{\pi(n)}\}.$$
Then $[0,1]^n = \bigcup\limits_{\pi \in S_n} C_\pi$, and for $\pi \neq \sigma$, the intersection $C_\pi \cap C_\sigma$ is a null set (it's lower-dimensional). Hence we have
$$\begin{align}
\left(\int_0^1 f(s)\,ds\right)^n &= \int_{[0,1]^n} \prod_{i=1}^n f(s_i) \,ds_1\dotsc ds_n\\
&= \sum_{\pi \in S_n} \int_{C_\pi} \prod_{i=1}^n f(s_i)\,ds_1\dotsc ds_n.
\end{align}$$
The product $\prod\limits_{i=1}^n f(s_i)$ is permutation-invariant, and all $C_\pi$ are congruent (permutation of the coordinates according to $\pi$ maps $C_{\operatorname{id}}$ to $C_\pi$), hence
$$\int_{C_\pi} \prod_{i=1}^n f(s_i)\,ds_1\dotsc ds_n$$
is independent of the permutation $\pi$.
Overall, that yields
$$\int_{C_\pi} \prod_{i=1}^n f(s_i)\,ds_1\dotsc ds_n = \frac{1}{n!}\left(\int_0^1 f(s)\,ds\right)^n$$
since there are $n!$ permutations of the $n$ coordinates.
