Answer
See below.
Work Step by Step
We know that ${n\choose r}=\frac{n!}{(n-r)!r!}$ by definition
Hence here: ${n\choose r}=\frac{n!}{(n-r)!r!}=\frac{n!}{r!(n-r)!}=\frac{n!}{(n-(n-r))!(n-r)!}={n\choose n-r}$.
Thus we proved what we wanted to.