Expected value of $\max\{X_1,\ldots,X_n\}$ where $X_i$ are iid uniform. Let $X_i\sim\mathrm{Uniform}(0,\theta)$ be iid, what is $E[\max\{X_1,\ldots,X_n\}]$? Apparently the answer is $$\frac{n}{n+1}\theta,$$
but I do not see why?
It seems intuitive in that you would "expect" them to be spaced out evenly, hence the maximum would be $\frac{n}{n+1}$-of-the-way along the interval, but how can we prove this mathematically? I feel like it should be simple, but evidently $$E[\max\{X_1,\ldots,X_n\}]\neq\max\{E[X_1],\ldots,E[X_n]\}.$$
Thanks.

Your intuition is correct.
To see this mathematically, suppose $X_1, \ldots, X_n$ are independent and uniformly distributed and  $M_n = \max(X_1,X_2,\ldots,X_n).$ 
The distribution function of the maximum is the joint probability that $X_k \leq x$ for all $k.$ This is a product of marginal probabilities since the variables are independent.
$$ F_M(x)=P(M_n \leq x) =P(X_1 \leq x,\ldots,X_n \leq x)=(x/\theta)^n$$
for $0 \leq x \leq \theta$. 
Also $F_M(x) = 0$ for $x < 0$ and $F_M(x) = 1$ for $x > \theta$.
Hence, the probability density function on $[0,\theta]$ is 
$$f_M(x)=F'_M(x)=nx^{n-1}\theta^{-n}$$
and the expected value is
$$E(M_n) = \theta^{-n}\int_0^{\theta}xnx^{n-1}\, dx=\frac{n}{n+1}\theta.$$
