Show that ${n+k-1 \choose k}={n+k-2\choose k}+{n+k-1 \choose k-1}$
So I have $${n+k-2\choose k}+{n+k-1 \choose k-1}=\frac{(n+k-2)!}{k!(n+k-2-k)!}+\frac{(n+k-1)!}{(k-1)!(n+k-1-k+1)!}$$
$$=\frac{n+k-2)!(k-1)!(n!)+(n+k-1)!(k!)(n-2)!}{(k!)(k-1)!(n-2)!(n)!}$$
$$=\frac{(n+k-1)!(k-1)!(n-2)![n(n+1)(n+k-1)+k]}{k!(n-2)!(k-1)!(n!)}$$
$$=\frac{(n+k-1)![n(n+1)(n+k-1)+k]}{k!(n!)}$$
From here I'm not sure how to continue. I want to end up with $\frac{(n+k-1)!}{k!(n+k-1-k)!}$
On
Here is a counterexample to show that the "equation" in your title is not true
(no wonder you were having trouble proving it!):
with $n=4$ and $k=2$, $$\binom52\ne\binom42+\binom51$$ since $10\ne6+5$.
You should have more success proving that $$\binom{n+k-1}{k}=\binom{n+k-2}{k}+\binom{n+k-\color{red}2}{k-1}.$$
In fact, $${n+k-1 \choose k}={n+k-2\choose k}+{n+k-2 \choose k-1}$$
This is because $\binom ak=\binom{a-1}k+\binom{a-1}{k-1}$ where $a=n+k-1$.