Mobius function sum for $\mu(k!)$

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Compute the sum $\sum\limits_{k=1}^n\mu(k!)$ for each $n \in \mathbb{N}$.

By definition:$$ \mu(n) = \begin{cases} 1; & n = 1\\ (-1)^r; & n = \prod\limits_{k = 1}^r p_k\\ 0; & \text{otherwise} \end{cases} $$ So if $k!=1\cdot2\cdot3\cdots(k-2)\cdot(k-1)\cdot k$, I would need to know the prime factorization of $k$ in order to know the value of $r$, correct?

So every other integer is odd, therefore $k!$ has $k/2$ integers which are not multiples of $2$. But where to go from here?

I know the final answer is $0$ since it is $0$ anytime $n\neq1$, but I think I need to show some sort of prime factorization on $k$ to get to the final answer.

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Hint: $\mu(n) = 0$ if $n$ is divisible by ...