First, your post has the frustration of stating the question in terms of $k$ balls and $n$ bins, and then your attempted answer is written in terms of $n$ balls and $k$ bins.
From here out, I will be referring to the question which has $n$ balls and $k$ bins with $n\geq k$.
In regards to the question of "Is my attempt correct?" the answer is no. You can see this with small examples such as what happens with $2$ balls and $1$ bin (as brought up in the comments above)
There is clearly only one way to distribute these balls into the one bin (all balls go into the only bin available). However, according to your method you incorrectly count two ways in which the balls go into the bin.
Your error is in forgetting that when applying multiplication principle, objects selected in earlier steps are treated as "different" than objects selected in later steps. For the case of two balls and one bin, the ball picked in the first step was treated as relevant when it shouldn't have been.
A counting problem which has the answer of $\binom{n}{k} k! k^{n-k}$ similar to the one you hoped would be something along the lines of:
You have $n$ red balls, each of which are distinctly numbered, and you have $k$ bins. In how many ways can you take distribute the balls among the bins such that every bin receives at least one ball and then from there take a ball from each bin and then color it gold.
(or equivalently, perform the actions in a different order: first pick $k$ balls, color them gold and place them in the bins, one gold ball per bin, and then place the rest of the balls in the bins)
The original intent of the problem does not have this additional step that we color balls gold before placing them in the bins. Any arrangement that we intended to count only once which has a bin with more than one ball in it was instead counted several times, once for each different choice of which ball was to be colored gold.
In regards to the question of "Can we perform a correct calculation while avoiding inclusion-exclusion principle?" the answer is yes.
One way is to brute force the count... breaking it up via cases as necessary. This is a terrible idea and becomes incredibly tedious very quickly.
Another way is somewhat cheating... but we can use Stirling numbers of the second kind in our calculations. I say this is cheating, because ordinarily one derives the formula for Stirling numbers of the second kind by using inclusion-exclusion principle in the first place. You can derive the formula using recurrence relations instead if desired.
We get that there are $\left\{\begin{smallmatrix} n\\k\end{smallmatrix}\right\}k!$ ways in which we can distribute the $n$ balls into the $k$ bins such that no bin is empty, giving a final probability as:
$$\dfrac{\left\{\begin{smallmatrix}n\\k\end{smallmatrix}\right\}k!}{k^n}$$
