We were giving the following equation to reduce multiple integrals ($n$-integrals) to single integrals $$ \int_0^x\int_0^x...\int_0^x(x-t)u(t)dtdt...dt=\frac{1}{n!}\int_0^x(x-t)^nu(t)dt $$
However, the integrals $\int_0^x\int_0^x...\int_0^x(x-t)u(t)dtdt...dt$ don't make sense to me because $\int_0^x(x-t)u(t)dt$ just gives a function of $x$, and integrating the other integrals just multiples $\int_0^x(x-t)u(t)dt$ by $x^{n-1}$. I tried some functions for $u(x)$ in Methematica and I found that LHS $\neq$ RHS.
We are using this textbook.
Too long for a comment, but not really a proper answer...
You say, "the integrals .. don't make sense to me... because it gives a function of $x$". You are 100% correct. The I guess, technically, $\int_0^x f(x) dt$ makes sense; $f(x)$ is constant wrt $t$, so the integral is equal to $x f(x)$---just $f(x)$ times the length of the range.
The only way I can see this working is if you replace the $dt$-s with $dx_1$, $dx_2$, ..., $dx_{n-1}$: $$ LHS = \int_0^t \int_0^{x} \cdots \int_0^{x_2} (x_1 - t) u(t) dx_1 \cdots dx_{n-1} dx dt.$$ This way, you keep on integrating $\int_0^{x_j} (x_{j-1} - t)^{j-1} dx_{j-1} = j^{-1} (x_j - t)^j$ and ignore the $u(t)$.
Edit. I see you've added a clip from the book. I don't know, sorry!