For $n = 0,1,2,...$ and $k \in \{0,...,n\}$, prove that $$\binom{n}{k+1}+\binom{n}{k}=\binom{n+1}{k}$$
Here's my attempt. $$\binom{n}{k+1}+\binom{n}{k}=\frac{n!}{(n-(k+1))!(k+1)!} + \frac{n!}{(n-k)!k!}$$ $$\implies \frac{n!(n-k)}{(n-(k+1))!(k+1)!(n-k)} + \frac{n!(k+1)}{(n-k)!k!(k+1)}$$ $$\implies \frac{n!(n-k)+n!(k+1)}{(n-k)!(k+1)!}$$ $$\implies \frac{n!((n-k)+(k+1))}{(n-k)!(k+1)!}$$ $$\implies \frac{n!(n+1)}{(n-k)!(k+1)!}$$ But I get stuck here because this does not equal $\binom{n+1}{k}$ but instead equals $\binom{n+1}{k+1}$.
Your proof is indeed correct as the identity itself is wrong. If you choose $k+1$ items out of $n+1$ items, you either choose all $k+1$ items from the first $n$ items, or choose $k$ items from the first $n$ items and choose the last item. Hence the right hand side must contain $k+1$ instead of $k$.