Problem 18-22 on p. 327 of Michael Spivak's Calculus (first edition) is
Use induction and integration by parts to show that $$\int_0^x \frac{f(u)(x-u)^n}{n!}du=\int_0^x \left( \int_0^{u_n}\left( \dotsb \left( \int_0^{u_1}f(t)\,dt \right) du_1 \right) \dotsb \right)du_n$$
Previous exercises (14-5 and 14-6) have asked us to prove essentially the same thing by induction and by noting that both sides have the same derivative with respect to $x$ and the same value at zero. So I can see how to do the problem that way.
When I try to solve by integration by parts, I'm getting something funny. The $n=1$ case works out OK. When I try to do $n=2$, for instance, let me show you what I'm getting. I want to show $$\int_0^x \frac{f(u)(x-u)^2}{2!}du = \int_0^x \int_0^{u_2}\int_0^{u_1}f(t)\,dt\,du_1\,du_2.$$ If I substitute into the right hand side, using the $n=1$ case, I get that it suffices to show
$$\int_0^x \frac{f(u)(x-u)^2}{2!}du =\int_0^x \left( \int_0^{u_2}f(u_1)(u_2-u_1)\,du_1 \right)du_2.$$ And if I integrate the LHS by parts, I get that it suffices to show
$$\int_0^x \int_0^u f(t)(x-u)\,dt\,du =\int_0^x \left( \int_0^{u_2}f(u_1)(u_2-u_1)\,du_1 \right)du_2.$$
But note these expressions are not the same.
I could always expand, etc. and show that it comes out right, but I'm trying to find the way by simple integration by parts (that is what the author wants me to see).
I know I'm just missing something simple!