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 $$F_{Y}(y) = P(Y \leq y) = \begin{cases} 0 & y \leq a \\ \phantom{} \left[ (y-a)/(b-a) \right]^n & y\in(a,b) \\ 1 & y \geq b \\ \end{cases}$$$$F_{Y}(y) = P(Y \leq y) = \begin{cases} 0 & y \leq a \\ \phantom{} \left( \frac{y-a}{b-a} \right)^n & y\in(a,b) \\ 1 & y \geq b \\ \end{cases}$$
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.
