5 cards are chosen from a standard deck. What is the probability that we get all four aces, plus the king of spades?

Assuming you just want those 5 cards in any order:

Since you have specified all 5 specific cards that you want, you don't need to consider the aces separately from the king. The probability of selecting these 5 cards is then

(Probability choosing any one of the 5) $\times$ (Probability of choosing one of the remaining 4) $\times$ ..., i.e. $$\frac{5}{52} \times \frac{4}{51} \times \frac{3}{50} \times \frac{2}{49} \times \frac{1}{48} = \frac{1}{54145} \times \frac{1}{48},$$ just to confirm what Ross Millikan said.

If, instead, you want the 4 aces before the king, we have $$\frac{4}{52} \times \frac{3}{51} \times \frac{2}{50} \times \frac{1}{49} \times \frac{1}{48}.$$

You've fully determined the five cards you want, so the probability is $\frac{1}{{52 \choose 5}}$ since there are ${52 \choose 5}$ different ways to choose 5 cards from the deck, and only one correct one.

Your first computation gets the chance you get four aces and any other card. The $48 \choose 1$ factor is selecting the other card. If you want specifically the king of spades it should be $1$, so the chance is $48$ times less. Your second computation gets the chance you draw the four aces first in some order, then draw the king of spades. You should multiply by $5$ for the number of positions the king can be in. Neither is correct for the question in the title.