Let $\left(x_i\right)_{i=1}^n$ be the sequence of $n$ random numbers generated and $\left(X_i\right)_{i=1}^n$ the iid random variables that generated them. For this problem, it is easier to look at $\mathbb{P}\left(X<x\right)$ for some $x\in\mathbb{R}$.
If we want to find $\mathbb{P}\left(X<x\right)$, $x\in\left(0,1\right)$ then we have the following:
\begin{align*}
\mathbb{P}\left(X<x\right) &= \mathbb{P}\left(X_1<x\cap\dotsb\cap X_n<x\right) \\
\text{due to independence}\qquad &= \prod_{i=1}^n\mathbb{P}\left(X_i<x\right) \\
\text{due to identical distribution}\qquad &= \prod_{i=1}^n\mathbb{P}\left(X<x\right) \\
\text{as each $X_i$ is uniform on $\left[0,1\right]$}\qquad &= x^n \\
\end{align*}
When $x\in\left(-\infty,0\right]$, $\mathbb{P}\left(X<x\right)=0$ as the maximum of the randomly generated numbers will always be greater than $0$. When we have $x\in\left[1,\infty\right)$, $\mathbb{P}\left(X<x\right)=1$ as the maximum of the randomly generated numbers will always be less than $1$.
So we have
$$
\mathbb{P}\left(X<x\right)=
\begin{cases}
0, & x\in\left(-\infty,0\right] \\
x^n, & x\in\left(0,1\right) \\
1, & x\in\left[1,\infty\right)
\end{cases}
$$
Differentiating this with respect to $x$ we obtain that
$$
\mathbb{P}\left(X=x\right)=
\begin{cases}
0, & x\in\left(-\infty,0\right] \\
nx^{n-1}, & x\in\left(0,1\right) \\
0, & x\in\left[1,\infty\right)
\end{cases}.
$$