<s>[INST] proving $\binom{n-1}{k} - \binom{n-1}{k-2} = \binom{n}{k} - \binom{n}{k-1} $ The identity is
$$\binom{n-1}{k} - \binom{n-1}{k-2} = \binom{n}{k} - \binom{n}{k-1}. $$
I wrote a matlab code and numerically checked it. It is right.
It is also not difficult to prove it by brute force. But can anyone come up with a simple proof using combinatorics? Many such identities are proven in this elegant way.
ps. It is an identity in the 'Group theory' book by Wigner.
 [/INST] Hint: You might also find appealing that algebraically this binomial identity is encoded by the relation
\begin{align*}
1-z^2=(1+z)(1-z)
\end{align*}
Denoting with $[z^k]$ the coefficient of $z^k$ of a series we can write
\begin{align*}
\binom{n}{k}=[z^k](1+z)^n\tag{1}
\end{align*}

Using (1) and noting that $[z^p]z^qA(z)=[z^{p-q}]A(z)$ we obtain
\begin{align*}
\color{blue}{\binom{n-1}{k}}\color{blue}{ - \binom{n-1}{k-2}}
&=[z^k](1+z)^{n-1}-[z^{k-2}](1+z)^{n-1}\\
&=[z^k]\left(1-z^2\right)(1+z)^{n-1}\\
&=[z^k](1-z)(1+z)^n\\
&=[z^k](1+z)^n-[z^{k-1}](1+z)^{n}\\
&\,\,\color{blue}{=\binom{n}{k}-\binom{n}{k-1}}
\end{align*}

</s>