This is a review problem for my final exam: Let $(X_{n})_{n\geq 1}$ be an i.i.d. sequence of random variables with $X_{i} \sim U(0,1)$. Let $M_{n}=\max_{1\leq i \leq n}X_{i}$. Show that $n(1-M_{n})$ converges weakly to an exponential distribution with parameter $1$.
I attempted to solve this problem using characteristic functions and Levy's Continuity Theorem. Perhaps the issue is what the characteristic function of $n(1-M_{n})$ is supposed to look like, because I was unable to get the expectation of that characteristic function to converge to $\frac{1}{1-iu}$, the characteristic function of an exponential with parameter $1$.
Anyway, this is what I did:
\begin{align} \varphi_{n(1-M_{n})}&=E\{exp(iun(1-M_{n}))\}=\exp(iun)E\{\exp(-iunM_{n})\}\\ &=\exp(iun)\cdot\frac{1}{1-0}\int_{0}^{1}\exp(-iunM_{n})dM\\ & =-\frac{1}{iun}[\exp(-iunM_{n})]_{0}^{1}\\ &=-\frac{1}{iun}[\exp(-iun)-1]=\frac{\exp(iun)+1}{iun}, \end{align} which does not go to $\frac{1}{1-iu}$ as $n \to \infty$.
What do I need to do differently in order to make it work? I want to be able to handle problems like this on the final, so please help!! If characteristic functions aren't necessarily the best way to approach this problem, please tell me a better way.
It's probably better to use the cumulative distribution function, since it behaves better with respect to the maximum of an independent sequence. We have $$\mu\{n(1-M_n)\leqslant t\}=\mu\{1-t/n\leqslant M_n\}=1-(1-t/n)^n$$ and we recognize a well-known limit.