By setting $N = 1 \cdot 2 \cdot 3 \cdot 4 \cdots 199 \cdot 200$ we obtain $N = \prod_{k=1}^{200} k! = \prod_{k=1}^{100} (2k-1)! \cdot (2k)! = \prod_{k=1}^{100} [(2k-1)!]^2 \cdot (2k) = \prod_{k=1}^{100} [(2k-1)!]^2 \cdot \prod_{k=1}^{100} 2 \cdot \prod_{k=1}^{100} k = \prod_{k=1}^{100} [(2k-1)!]^2 \cdot 2^{100} \cdot 100!$, yielding $\frac{N}{100!} = \big( 2^{50} \cdot \prod_{k=1}^{100} (2k-1)! \big)^2$. Hence, since ${\textstyle \frac{N}{100!}}$ is a perfect square, the number Julia erased was $100!$.