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

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