I know that when calculating the probability of one result, e.g., getting at least one 4, you would use the equation -
$1-(3/4)^9$
But I'm not sure what kind of equation you would use to find out the probability of getting at least one of each.
By one of each value I mean a result that has at least one 1, at least one 2, at least one 3 and at least one 4.
For example, I'm looking for a result like, (1, 2, 3, 4, 4, 4, 4, 4, 4) Or (1, 2, 3, 3, 4, 4, 4, 4, 4), etc...
Something that would not meet the criteria is (1,1,1,2,2,2,3,3,3)
One way of reading your expression is that there are $4^9$ ordered ways for the dice to fall, of which $3^9$ have only $1,2,3$ showing, so the chance you get at least one $4$ is $\frac {4^9-3^9}{4^9}$. You have subtracted the throws without a $4$ from all the throws.
It would seem then the chance you are not missing any number is $\frac {4^9-4\cdot 3^9}{4^9}$ because you have four choices of the number to be missing. The problem is that you have subtracted rolls with two numbers missing twice and rolls with three numbers missing three times. We add in the rolls with only two numbers, of which there are ${4 \choose 2}2^9$ but now we have added back the ones with only one number three times, so we need to subtract them once. The final answer is $$\frac {4^9-{4 \choose 1}3^9+{4 \choose 2}2^9-{4\choose 3}1^9}{4^9}$$ This is an example of the inclusion-exclusion principle