I want to show that : $$\binom{n}{k}+\binom{n}{k-1} = \frac{(n+1)!}{k!(n+1-k)!}$$
Here is my proof : $\forall 1\leq k\leq n$ :
$$\begin{align} \binom{n}{k}+\binom{n}{k-1} &= \frac{n!}{k!(n-k)!} + \frac{n!}{(k-1)!(n+1-k)!} \tag1 \\[4pt] &=\frac{n!(n+1-k)}{k!(n+1-k)!} +\frac{n!k}{k!(n+1-k)!} \tag2 \\[4pt] &= \frac{n!(n+1-k+k)}{k!(n+1-k)!} \tag3 \\[4pt] &=\frac{(n+1)!}{k!(n+1-k)!} \tag4 \\[4pt] &=\binom{n+1}{k} \tag5 \end{align}$$
And normally I avoided problem of factorial not defined since $k\leq n$.
Do you think this is correct ?
Thank you a lot
EDIT : Thank you everyone and especially Riemann and Jean-Claude Arbault since it is not defined for $k=0$...
Yes it's correct $$\binom{n}k+\binom{n}{k-1}$$ $$=\frac{n!}{(n-k)!\cdot k!}+\frac{n!}{(n+1-k)!\cdot (k-1)!}$$ $$=\frac{n!}{(n-k)!\cdot (k-1)!}\left(\frac{1}{k}+\frac{1}{n+1-k}\right)$$ $$=\frac{n!}{(n-k)!\cdot (k-1)!}\left(\frac{n+1-k+k}{k(n+1-k)}\right)$$ $$=\frac{n!(n+1)}{(n+1-k)!\cdot k!}$$ $$=\frac{(n+1)!}{(n+1-k)!\cdot k!}$$ $$=\binom{n+1}{k}$$ Here we made the assumption that $k\ne 0$