I decided to prove this by induction and I'm not sure if this is a valid proof, or maybe there is something that I'm missing. Anyway, I'd appreciate some feedback nonetheless.
Note: This proof assumes knowledge of the Möbius function.
Proof that $\sum_{k=1}^{n}\mu(k!)=1$ for $n \geq 3$:
Base Case
- Let $n=3$ and test:
$\sum_{k=1}^{3}\mu(k!) = \mu(1) + \mu(2 \cdot 1) + \mu(3 \cdot 2 \cdot 1)$
$=(1)+(-1)+(1)=1$
- True for base case ($n=3$).
Induction Hypothesis
- Assume that it is true for $m$ for $m \geq 3$:
$\sum_{k=1}^{m}\mu(k!)=1$
Inductive Step:
- Using the Inductive Hypothesis as a premise, we have:
$\sum_{k=1}^{m+1}\mu(k!)=(\sum_{k=1}^{m}\mu(k!)=1)+\mu((m+1)!)$
$=1+\mu((m+1)!)$
Since $m\geq 3$, we have $m+1\geq 4$. But then $(m+1)!=(m+1)(m)...(4)(3)(2)(1)$ Notice the factor of $(4)$ in the expanded sum.
Since $4=2^2$ is a square number, $(m+1)!$ has a squared factor, and according to the definition of the Möbius function, $\mu((m+1)!)=0$.
Finally, we have
$=1+\mu((m+1)!) = 1+0 = 1$
- True for inductive step, therefore true for all cases.