I know this question has been asked before both here and here but my question revolves around why my own attempt was incorrect and if someone could help guide me into why we need Sterling numbers.
The total number of all possible selections is straightforward $k^n$, which will be our denominator.
For our numerator, I said first select $k$ objects from our $n$ selected that we know must contain all $k$ groups at least once. This is $\binom{n}{k}$, then you can say that the remaining objects have $k^{n-k}$ possibilities.
So $$ \frac{\binom{n}{k}k^{n-k}}{k^n} = \frac{\binom{n}{k}}{k^k} $$
which makes no sense as this grows unbounded for fixed $k$. Any assistance into where my error could be would be very appreciated!
Your attempt is incorrect because you have fractionated the selection by first choosing one of each item, and then choosing the remaining items any which way. But this means that you are choosing an unknown number of items already chosen an unknown number of times, and thus overcounting
If you want to avoid Stirling numbers,(e.g. you might not have easy recourse to a table of Stirling numbers) a simple way is to use inclusion-exclusion. Suppose you want to put 6 distinct balls into 4 distinct boxes with none empty.
Number of ways = All ways - at least one box empty + at least two boxes empty - ....
$and\; Pr = \left[4^6 - \binom41 3^6 + \binom 42 2^6 - \binom43 1^6\right]\Big /4^6$
You should be able to transform this for $n$ items from $k$ groups