# why is $\frac{k\cdot n!}{k!(n-k)!} = \frac{n(n-1)!}{(k-1)!((n-1)-(k-1))!}$?

This is an equation from my textbook that I am trying to understand:

$$\frac{k\cdot n!}{k!(n-k)!} = \frac{n(n-1)!}{(k-1)!((n-1)-(k-1))!}$$

What I got so far, is that $$\frac{k\cdot n!}{k!} = \frac{n!}{(k-1)!}$$ however, why does the same principle apply for (n-k)! in the denominator? Isn't there only one k in the numerator that I can cancel out in the denominator?

In other words, shouldn't it be $$\frac{k\cdot n!}{k!(n-k)!} = \frac{n(n-1)!}{(k-1)!((n-k)!}$$ ?

Hint: Simple! use the following simple equations: $$(n-1) - (k-1) = n - k$$ And: $$n! = n \times (n-1)!$$
All you have to do is to rewrite $$\dfrac{k}{k!}$$ as $$\dfrac{1}{(k-1)!}$$ and note that $$n! = n(n-1)!$$ and $$(n-1)-(k-1) = n-k$$.