Since this is a graduate level number theory class, I think it's safe to assume that you are familiar with modulo arithmetic?
Given any list of $n$ consecutive integers, $a, a+1, a+2, \dots, a+n-1$, modulo $n$ this list is equivalent to $0,1,2,3,\dots,n-1$ modulo $n$. (Note that I am not saying $a \equiv 0 \pmod{n}$). This list can be rewritten as:
$1 \equiv 1 \pmod{n}$
$2 \equiv 2 \pmod{n}$
$3 \equiv 3 \pmod{n}$
$\dots$
$\dfrac{(n-1)}{2} \equiv \dfrac{(n-1)}{2} \pmod{n}$
$n-1 \equiv -1 \pmod{n}$
$n-2 \equiv -2 \pmod{n}$
$n-3 \equiv -3 \pmod{n}$
$\dots$
$\dfrac{(n+1)}{2} \equiv -\dfrac{(n-1)}{2} \pmod{n}$
Since $n$ is odd, exponentiation preserves the sign. And so
$$0^n + 1^n + 2^n + \dots + \left(\dfrac{n-1}{2}\right)^n + \left(\dfrac{n+1}{2}\right)^n + \dots + (n-2)^n + (n-1)^n + n^n$$
is equivalent to
$$1^n + 2^n + \dots + \left(\dfrac{n-1}{2}\right)^n - \left(\dfrac{n-1}{2}\right)^n + \dots - 2^n - 1^n$$
modulo $n$, and so the sum becomes $0$ modulo $n$. Note that the exponent could be replaced by any odd integer and the statement will still hold.
EDIT: Here is the example you requested in the comments.
Let's say we have the list 5, 6, 7, 8, 9, 10, 11, with $n=7$.
Okay, so the first thing I will do is to find their representatives in $\mod 7$ between $0$ and $6$ inclusive.
So,
5, 6, 7, 8, 9, 10, 11 becomes 5, 6, 0, 1, 2, 3, 4.
Now, let's look at $(n-1)/2$. For $n=7$, this number is 3. That is the cut off for the positive terms. The rest of them I turn them into negatives:
$5, 6, 7, 8, 9, 10, 11$ becomes
$5, 6, 0, 1, 2, 3, 4,$ which is
$7-2, 7-1, 0, 1, 2, 3, 7-3$, which becomes
$-2, -1, 0, 1, 2, 3, -3$.
Now, take any odd power of these numbers, sum, you get 0 in modulo 7.
Technically, I could've gone directly from the original list to $-2, -1, 0, 1, 2, 3, -3$, without going through the intermediate step, but I wanted to illustrate how the proof applies to this particular example.
Best Answer
This is an application of the Pigeonhole Principle. The idea is that since there are $n$ possible remainders when an integer is divided by $n$ that at least two of the $n + 1$ integers in the set $\{x_0, x_1, \ldots, x_n\}$ must have the same remainder when divided by $n$. If they have the same remainder when divided by $n$, their difference is divisible by $n$.
The possible remainders when an integer is divided by $n$ are $0, 1, 2, \ldots, n - 1$. If you have $n$ integers, then it is possible for each of them to have a different remainder when divided by $n$. However, if you have $n + 1$ integers, at least two of them must have the same remainder when divided by $n$. Hence, in the set $\{x_0, x_1, \ldots, x_n\}$, there are two numbers $x_i$ and $x_j$, with $i \neq j$, that have the same remainder when they are divided by $n$. Thus, there exist $k, m, r \in \mathbb{N}$, with $0 \leq r \leq n - 1$, such that
\begin{align*} x_i & = kn + r\\ x_j & = mn + r \end{align*}
If we take their difference, we obtain
$$x_i - x_j = (kn + r) - (mn + r) = kn - mn = (k - m)n$$
Therefore, $x_i - x_j$ is divisible by $n$.