Proving $\eqalign{n\choose k} = {n\choose k-1} \frac{n-k+1}{k}$

Question

I want to prove that

$$\eqalign{n\choose k} = {n\choose k-1} \frac{n-k+1}{k}$$

So I can just write

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

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

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

I think I did a mistake in the last line, because I cancelled $(n-k+1)!$ since $(n-k+1)! = (n-k+1) \cdot (n-k)!$

So, is my mistake that I cancelled faculty wrongly?

Answer 1

Your mistake is you forgot the $k$ on the denominator of the rightmost fraction, and somehow dropped the factorial sign on the $(k-1)$ in the final fraction.

Note that $k \cdot (k-1)! = k!$ and you will have verified the identity.

Answer 2

$$ \frac {n!}{(n−k)!⋅k!} = \frac {n!(n-k+1)}{(n−k+1)!⋅(k−1)!k} $$

At this point you are done $$ = \frac {n!}{(n−k)!⋅k!}$$