Convergence of uniform, independent RV's

$X_1, X_2,.....$ are independent random variables that all have a uniform distribution, U(0,1).

Prove that if $M_n = max (X_1, X_2, ..., X_n)$ then as n approaches infinity, $n(1-M_n)$ converges in distribution to an Exp(1) random variable.

I've done a little bit but I have confused myself and I don't know what to do with it now (and im not even sure if Im on the right track (Worried)).

So

$Pr(n(1-M_n) \leq x) = Pr(1 - \frac{x}{n} \leq M_n)$

$= 1 - Pr(M_n < 1 - \frac{x}{n})$

$= 1 - Pr(\forall n, X_n < 1 - \frac{x}{n})$

$= 1 - \prod_{i=1}^n \ Pr(X_i < 1 - \frac{x}{n})$

$= 1 - (1-\frac{x}{n})^n$

and this is where I get lost. I think maybe.....

$= 1 - exp(n \ln (1 - \frac{x}{n}))$

but then where to go from here I dont know, which makes me doubt the accuracy of what I have done. Thanks for any help you can provide me with.

you did the hard work, you just need to review calc 2

$\biggl(1-\frac{x}{n}\biggr)^n\to e^{-x}$ done

Ahhh, off course. (Headbang) Why I overcomplicate things sometimes I'll never know (Giggle)

Don't be so hard on yourself.
We have Moo at this forum for that. (Smirk)