How do you calculate the probability density function of the maximum of a sample of IID uniform random variables? Given the random variable
$$Y = \max(X_1, X_2, \ldots, X_n)$$
where $X_i$ are IID uniform variables, how do I calculate the PDF of $Y$?
 A: If $F_{Y}(y)$ is the CDF of $Y$, then 
$$F_Y(y)=\text{Prob}(y>X_1,y>X_2,...,y>X_n)$$
You can then use the iid property and the cdf of a uniform variate to compute $F_Y(y)$.
A: The maximum of a sample is one of the order statistics, in particular the $n$th order statistic of the sample $X_1,\dots,X_n$. In general, computing the distribution of order statistics is difficult, as described by the Wikipedia article; for some special distributions, the order statistics are well-known (e.g. for the uniform distribution, which has Beta-distributed order statistics).
EDIT: The Wikipedia article on sample maximum and minimum is also helpful and more specific to your problem.
A: It is possible that this question is homework but I felt this classical elementary probability question was still lacking a complete answer after several months, so I'll give one here.
From the problem statement, we want the distribution of
$$Y = \max \{ X_1, ..., X_n \}$$
where $X_1, ..., X_n$ are iid ${\rm Uniform}(a,b)$. We know that $Y < x$ if and only if every element of the sample is less than $x$. Then this, as indicated in @varty's hint, combined with the fact that the $X_i$'s are independent, allows us to deduce
$$ P(Y \leq x) = P(X_1 \leq x, ..., X_n \leq x) = \prod_{i=1}^{n} P(X_i \leq x) = F_{X}(x)^n$$
where $F_{X}(x)$ is the CDF of the uniform distribution that is $\frac{y-a}{b-a}$. Therefore the CDF of $Y$ is
$$F_{Y}(y) = P(Y \leq y) = \begin{cases} 
0 & y \leq a \\ 
\phantom{} \left( \frac{y-a}{b-a} \right)^n & y\in(a,b) \\
1 & y \geq b \\ 
\end{cases}$$
Since $Y$ has an absolutely continuous distribution we can derive its density by differentiating the CDF. Therefore the density of $Y$ is
$$ p_{Y}(y) = \frac{n(y-a)^{n-1}}{(b-a)^{n}}$$
In the special case where $a=0,b=1$, we have that $p_{Y}(y)=ny^{n-1}$, which is the density of a Beta distribution with $\alpha=n$ and $\beta=1$, since ${\rm Beta}(n,1) = \frac{\Gamma(n+1)}{\Gamma(n)\Gamma(1)}=\frac{n!}{(n-1)!} = n$.
As a note, the sequence you get if you were to sort your sample in increasing order - $X_{(1)}, ..., X_{(n)}$ - are called the order statistics. A generalization of this answer is that all order statistics of a ${\rm Uniform}(0,1)$ distributed sample have a Beta distribution, as noted in @bnaul's answer.
A: The maximum of a set of IID random variables when appropriately normalized will generally converge to one of the three extreme value types.  This is Gnedenko's theorem,the equivalence of the central limit theorem for extremes.  The particular type depends on the tail behavior of the population distribution.  Knowing this you can use the limiting distribution to approximate the distribution for the maximum.
Since the uniform distribution on [a, b] is the subject of this question Macro has given the exact distribution for any n and a very nice answer.  The result is rather trivial.  For the normal distribution a nice closed form is not possible but appropriately normalized the maximum for the normal converges to the Gumbel distribution F(x)=exp(- e $^-$$^x$).
For the uniform the normalization is (b-a)-x/n and F$^n$(b-a-x/n)=(1-x/[n(b-a)])$^n$ 
which converges to e$^-$$^x$$^/$$^($$^b$$^-$$^a$$^)$.  Note here that y=b-a-x/n.  and F$^n$(y) converges to 1 as y goes to b-a. This holds for all 0

In this case it is easy to compare the exact value to its asymptotic limit.
Gumbel's book
Galambos' book
Leadbetter's book
Novak's book
Coles book
