Algebraic proof regarding falling factorials 
Let $n$ and $k$ be positive integers with $n \ge k$.
Give an algebraic proof that $$(n)_k = (n-1)_k + k\cdot(n-1)_{k-1},$$ where $(n)_k$ is a falling factorial: $(n)_k$ = $(n)(n-1)(n-2)\ldots(n-k+1)$.

I started by turning LHS into
$$\frac{n!}{(n-k)!} = \frac{((n+k)-k)!}{(n-k)!}$$
I tried doing this after:
$$ = \frac{(((n+k)-k)(n-1))!}{(n-k)!}$$
but I really have no idea where else to proceed. I'm not sure how I can split it up or move past this.
Does anyone think they can point me in the right path?

$(n)_k-(n-1)_k=\frac{n!}{(n-k)!}-\frac{(n-1)!}{(n-k-1)!}=\frac{(n-1)!}{(n-k-1)!}(\frac{n}{n-k}-1)=\frac{(n-1)!}{(n-k-1)!}\times \frac{k}{n-k}=\frac{(n-1)!}{(n-k)!}\times k=k(n-1)_{k-1}$
