Suppose there are $n + 1$ students and $n$ teaching assistants (T.A.'s). So how many ways can we assign the students to T.A.'s when each T.A. can have more than $1$ student?
When I did it, I thought about it as a combinations problem and got $C(2n, n)$, but I am not sure if that is the answer.
Since students and teaching assistants are people, they are distinguishable. Since there are $n$ possible teaching assistants to which each student could be assigned, there are $n^{n + 1}$ possible assignments of teaching assistants to students.
It appears that you ignored the fact that students are distinguishable. If we were only interested in how many students are assigned to each teaching assistant, then the number of possible assignments is the number of solutions of the equation $$x_1 + x_2 + x_3 + \cdots + x_n = n + 1$$ in the nonnegative integers, where $x_j$ is the number of students assigned to the $j$th teaching assistant. Since a particular solution of that equation corresponds to the placement of $n - 1$ addition signs in a row of $n + 1$ ones, there are $$\binom{n + 1 + n - 1}{n - 1} = \binom{2n}{n - 1}$$ ways to do that because we must choose which $n - 1$ of the $2n$ positions required for $n + 1$ ones and $n - 1$ addition signs will be filled with addition signs.