I would like to prove the following identity:
$$ \int_{0}^{t} \left [\frac{(t - s)^n}{n!}f(s) - \int_{0}^{s} \frac{(s - x)^{n - 1}}{(n - 1)!} f(x) dx \right] ds = 0. $$
Yes, I know it's really ugly. My motivation for caring about the thing is that I was writing a proof about the spectral radius of an operator, and this is precisely what I need to be true for the inductive step to hold. So far, I've tried simply evaluating it with specific functions $f$, and I've tried expanding the $(t - s)^n$ and $(s - x)^{n - 1}$ terms (with the binomial theorem) in order to get the sums of some simpler looking integrals. That approach doesn't seem to work, and I'm not sure how else to tackle it given a general $f$. I get the sense there's some easy way to verify this that I'm missing; I'd appreciate any ideas!
You can use the Cauchy formula for repeated integration
$$ \int_{0}^{t} \frac{(t - s)^n}{n!}f(s) ds = f^{(-n)}(t) $$
$$ \int_{0}^{t} \left [\int_{0}^{s} \frac{(s - x)^{n - 1}}{(n - 1)!} f(x) dx \right] ds = \int_{0}^{t} f^{(-(n-1))}(s) ds = \int_{0}^{t} \left((f^{(-n)}\right)'(s) ds=f^{(-n)}(t) $$
Hence, we deduce your identity.