Algebraically, $$\dfrac{n!}{(n-r)!\;r!}=\dfrac{(n-1)!}{(n-r)!(r-1)!}+\dfrac{(n-1)!}{(n-1-r)!\;r!}$$ is equivalent to $$n!=r(n-1)!+(n-1)!(n-r)$$ on multiplying all terms by $r!(n-r)!$. Dividing through by $(n-1)!$ gives $$n=r+(n-r) \, ,$$ i.e., $n = n$.

Algebraically, $$\dfrac{n!}{(n-r)!\;r!}=\dfrac{(n-1)!}{(n-r)!(r-1)!}+\dfrac{(n-1)!}{(n-1-r)!\;r!}$$ is equivalent to $$n!=r(n-1)!+(n-1)!(n-r)$$ on multiplying all terms by $r!(n-r)!$. Dividing through by $(n-1)!$ gives $$n=r+(n-r) \, ,$$ i.e., $n = n$.

QED

A non-combinatorial way:

$$\dfrac{n!}{(n-r)!\;r!}=\dfrac{(n-1)!}{(n-r)!(r-1)!}+\dfrac{(n-1)!}{(n-1-r)!\;r!}$$ Multiplying all terms by $r!(n-r)!$ $$n!=r(n-1)!+(n-1)!(n-r)$$ Dividing everything by $(n-1)!$ $$n=r+(n-r)$$ $$n=n$$ QED

Algebraically, $$\dfrac{n!}{(n-r)!\;r!}=\dfrac{(n-1)!}{(n-r)!(r-1)!}+\dfrac{(n-1)!}{(n-1-r)!\;r!}$$ is equivalent to $$n!=r(n-1)!+(n-1)!(n-r)$$ on multiplying all terms by $r!(n-r)!$. Dividing through by $(n-1)!$ gives $$n=r+(n-r) \, ,$$ i.e., $n = n$.