<s>[INST] Proof that $\binom{n + 3}{4} = n + 3 \binom{n + 2}{4} - 3 \binom{n + 1}{4} + \binom{n}{4}$. I was trying to count the number of equilateral triangles with vertices in an regular triangular array of points with n rows. After putting the first few rows into OEIS, I saw that this was described by A000332: $\binom{n}{4} = n(n-1)(n-2)(n-3)/24$. 
A000332 has the comment:

Also the number of equilateral triangles with vertices in an equilateral triangular array of points with n rows (offset 1), with any orientation.

The linked solution is insightful, but the proof is very mechanical.
I tried to write an inductive proof, but I'm unable to come up with a nice way to prove the final identity (in particular for $n \geq 4$): 
$$\binom{n + 3}{4} = n + 3 \binom{n + 2}{4} - 3 \binom{n + 1}{4} + \binom{n}{4}$$.
Wolfram|Alpha admits that these are indeed equal, and this could certainly be shown by writing everything as a polynomial.
However, I was hoping to find some nice combinatorial identities that give some intuition as to why this equality holds.
 [/INST] $$\begin{align}
\binom {n+3}4-3\binom {n+2}4+3\binom {n+1}4-\binom n4
&=\sum_{r=0}^3(-1)^{r+1}\binom 3r\binom {n+r}4\\
&=\sum_{r=0}^3(-1)^{r+1}\binom 3r\binom {n+r}{n+r-4}\\
&=\sum_{r=0}^3(-1)^{r+1}\binom 3r (-1)^{n+r-4}\binom {-5}{n+r-4}\tag{*}\\
&=(-1)^{n-3}\sum_{r=0}^3\binom 3{3-r}\binom {-5}{n+r-4}\\
&=(-1)^{n-1}\binom {-2}{n-1}\tag{**}\\
&=(-1)^{n-1}\cdot (-1)^{n-1}\binom n{n-1}\tag{*}\\
&=\binom n1\\
&=n\\\\
\binom {n+3}4&=n+3\binom {n+2}4-3\binom {n+1}4+\binom n4\;\color{red}\blacksquare
\end{align}$$
*using upper negation
**using the Vandermonde Identity
</s>