I successfully completed part (ii), driving at the desired result of $I_{n+1} = -\frac{n+1}{m+1}I_n$. Using this recurrence relation and the fact that $I_0 = \frac{1}{m+1}$, I calculated $$I_n = (-1)^n \frac{(n+1)!}{(m+1)^{n+1}}$$
For part (iii), I wrote $x^x$ as $e^{x\log(x)}$, which I then wrote as the power series $\sum_{n=0}^{\infty} \frac{(x\log(x))^n}{n!}$. Then I swapped the integral and the sum, which led to me integrating $\frac{(x\log(x))^n}{n!}$. Given the result from part (ii), this was really easy.