This problem is from Ch. 5, Mathematical Statistics and Data Analysis, 3rd edition, stated as follows,
In addition to limit theorems that deal with sums, there are limit theorems that deal with extreme values such as maxima or minima. Here is an example. Let $U_1, \ldots, U_n$ be independent uniform random variables on $[0,1]$, and let $U_{(n)}$ be the maximum. Find the cdf of $U_{(n)}$ and a standardized $U_{(n)}$, and show that the cdf of the standardized variable tends to a limiting value.
Below is my derivation. The cdf of $U_{(n)}$ is as follows, $$ F_{(n)} = \left\{ \begin{array}{lr} x^n, & \text{if } x \in [0,1]\\ 0, & \text{otherwise} \end{array} \right\} $$ and pdf $$ f_{(n)} = \left\{ \begin{array}{lr} nx^{n-1}, & \text{if } x \in [0,1]\\ 0, & \text{otherwise} \end{array} \right\} $$ With pdf we can calculate mean and variance, which is $$ E(U_{(n)}) = \frac{n}{n+1}, \text{Var}(U_{(n)}) = \frac{n}{(n+1)^2(n+2)} $$ The standardized variable $Z_n$ is $$ Z_{n} = \frac{U_{(n)}-\frac{n}{n+1}}{\sqrt{\frac{n}{(n+1)^2(n+2)}}} $$
However, the key is quite different from mine. The standardized variable it gives is $Z_n = n(U_{(n)}-1)$. Did I miss something? Many thanks!
$U_{(n)}$ has a Beta$(n,1)$ distribution, sometimes called a standard power function. This gives the mean and variance you have stated.
The distribution of $n(1-U_{(n)})$ converges in distribution to an exponential distribution with rate $1$ so with mean and variance and standard deviation each $1$, and thus $n(U_{(n)}-1)$ converges in distribution to its negative.
Relocating this to have mean $0$ is possible but a slightly unusual thing to do for an exponential distribution. If you did, you would in effect be saying that the distribution of the standardised value of $U_{(n)}$ you stated has a limiting distribution with CDF $$F(x) = \min(e^{x-1},1)$$ the same as $1-Y$ where $Y\sim Exp(1)$.