How many ways are there to assign 24 students to five faculty advisors? Sorry if this seems trivial. I'm starting to think I'm pretty stupid for not understanding such a simple problem.
There are no other restrictions. An advisor may get multiple students, but one student may not get multiple advisors. 
I approached this by drawing $5$ slots corresponding to advisors and figuring out how many students can be assigned, without repetition (i.e. once a student gets assigned, they are no longer reused):
[24] x [23] x [22] x [21] x [20] = $P(24, 5)$
The book says that I should actually be treating the students as the slots and assigning instructors:
[5] x [5] x [5] x [5] x .... = $5^{24}$ 
Why is my logic wrong?
 A: Your approach only assigns one student to each adviser.  The other $19$ students do not get an adviser at all.  The question asks you to send each student to an adviser, with the possibility that one adviser gets all the students.  They are two very different questions with very different answers.
A: "Assigning" might make you think of functions, which assign an output value for each input value.
The number of functions 
$f: \{1,2,3,\dots ,n \} \to \{1,2,3,\dots ,m \}$ 
is equal to $m^n$.
A: It all comes down to it being functional (one input, one output). You can make a Cartesian product of the sets of students, and the set of advisors. there will be 120=24(5) placement options for the relation is placed with. If you color, or don't, then there's $2^{120}$ possible relations without restriction. but that includes one student being with all 5 advisors. that's not possible ( by the rules), so how many ways can one student choose exactly 1 advisor from the 5, $\binom 51=5$.There are 24 students. Each that can make that choice, and their choices are assumed independent. If so the number of possibilities for each is multiplied together. Part of the reason, your math is off, is because it doesn't take the fact that at least one advisor must have less than than 5 students ( see the pidgeonhole principle for reason why). Another reason, is that you calculated the number of ways each advisor, could pick 5 students, if order was important (at least one doesn't even have that many possibilities, and 2 order doesn't matter in theory so divide by 5! for the number of orders of each 5). etc.
