This is a variant of Xpw's answer, intended to reduce the number of separate cases that need to be examined in order to show that $(a,b,c)=(100,99,10)$ minimizes the sum $a+b+c$ if $c\ge10$.
Suppose $a!=b!\cdot c^2$ with $c\ge10$ and $a+b+c\lt209$. Let's write $a=b+n$. Then we have
We cannot have $n=2$, since any two consecutive integers are relatively prime, hence each would have to be a square in order for their product to be a square, but positive squares differ by at least $3$.
For $n\ge3$ we have the following crude inequality:
(You can be less crude if you like, but there's little to be gained from it.) Since $40^3=64000$ and $20^4=160000$, we see that $b\lt40$ if $n=3$ and $b\lt20$ if $n\ge4$. Let's do the $n\ge4$ case first.
Any string of $4$ or more consecutive numbers starting at a number less than or equal to $20$ contains at least one prime within the final $4$ numbers in the string (the first prime gap greater than $4$ occurs at $29-23$). The product of such a string contains that prime to the first power only, hence cannot be a square.
Any string of three consecutive numbers contains exactly one number divisible by $3$. In order for their product to be a square, the number divisible by $3$ must be divisible by an even power of $3$. For strings beginning at $40$ or less, the only possibilities are $9$, $18$, and $36$. Since $7$, $11$, $17$, $19$, and $37$ are primes, the only strings to check are $8\cdot9\cdot10$ and $34\cdot35\cdot36$, neither of which is a square.