Let $f(x)=\displaystyle\int_0^x \displaystyle\sum_{i=0}^{n-1} \dfrac{(x-t)^i}{i!}dt.$ Find the nth derivative $f^{(n)}x.$

Substitute $t =x-s$ in the integral: $$ f(x)=\int_0^x \sum_{i=0}^{n-1} \dfrac{(x-t)^i}{i!}dt = \int_0^x \sum_{i=0}^{n-1} \dfrac{s^i}{i!}ds $$ so that the integrand does not depend on $x$ anymore. Now differentiation becomes simple, e.g.: $$ f'(x) = \sum_{i=0}^{n-1} \dfrac{x^i}{i!} $$ etc.