It is possible to "cut short" a factorial expression by using the following equality: $$$n! = n \cdot (n-1)!$$$

This allows us to simplify terms when factorials appear in fractions.

For instance, calculating an expression like: $$$\dfrac{8!}{6!\cdot3!}$$$ We must bear in mind that in the numerator $$8! = 8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1 = 8\cdot7\cdot6!$$ (we have stopped the development at $$6!$$ because it is the term that appears in the denominator and thus we will be able to simplify it. So that $$$\dfrac{8!}{6!\cdot3!}=\dfrac{8\cdot7\cdot\cancel{6!}}{\cancel{6!}\cdot3!}= \dfrac{8\cdot7}{3\cdot2}=\dfrac{28}{3}$$$

Another example, calculating the value of: $$$\dfrac{14!\cdot6!}{13!\cdot7!}$$$ In this case we will develop the numerator and denominator in such a way as to have the biggest advantage in the simplification, $$14! = 14 \cdot 13!$$ and $$7! = 7\cdot6!$$:

$$$\dfrac{14!\cdot6!}{13!\cdot7!}=\dfrac{14\cdot \cancel{13!}\cdot \cancel{6!}}{\cancel{13!}\cdot7\cdot\cancel{6!}}=\dfrac{14}{7}=2$$$

The same method can be used for literal expressions (those in which letters appear instead of numbers): $$$\dfrac{x!}{(x-1)!}=\dfrac{x\cdot \cancel{(x-1)!}}{\cancel{(x-1)!}}=x $$$

The example can be as complicated as you like, but the solution will always be simple: $$$\dfrac{(m-2)!\cdot x!}{(x-1)!\cdot m!}=\dfrac{(m-2)!\cdot x\cdot\cancel{(x-1)!}}{\cancel{(x-1)!}m\cdot (m-1)!}= \dfrac{\cancel{(m-2)!}\cdot x}{m\cdot(m-1)\cdot\cancel{(m-2)!}}= \dfrac{x}{m\cdot(m-1)} $$$