I have a bowl of $n$ balls numbered $1$ to $n$. I will perform $k$ extractions (with immediate replacement into the bowl after extraction). I want to calculate the probability to have extracted each ball at least once after those $k$ turns.
With inclusion/exclusion principle from Probability to see all 6 numbers on a die after n throws for example, I arrived at:
$$P(n, k) = \sum_{i=1}^{n} (-1)^{n-i} \binom{n}{i} i^k \frac{1}{n^{k}}$$
But this takes forever to calculate for something like $n=50$, $k=1000$. Can I further simplify this equation? Is the equation even correct?
Your formula is correct. It can't be further simplified, except perhaps by expressing it in terms of the Stirling numbers of the second kind:
$$ \def\stir#1#2{\left\{#1\atop#2\right\}} P(n,k)=\frac{k!}{n^k}\stir kn\;. $$
There are other ways to calculate this probability (for instance by keeping track of the probability distribution over the number of distinct balls seen and updating it $k$ times), but I'd be surprised if any of them took less than the $O(kn)$ steps that evaluating your equation takes.