<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] This is calculus of finite differences:
$$\binom{n+1}2-\binom{n}2=\binom{n}1=n,$$
$$\binom{n+1}3-\binom{n}3=\binom{n}2,$$
$$\binom{n+1}4-\binom{n}4=\binom{n}3,$$
etc. Then
$$n=\binom{n+1}2-\binom{n}2
=\left(\binom{n+2}3-\binom{n+1}3\right)
-\left(\binom{n+1}3-\binom{n}3\right)
=\binom{n+2}3-2\binom{n+1}3+\binom{n}3
=\left(\binom{n+3}4-\binom{n+2}4\right)
-2\left(\binom{n+2}4-\binom{n+1}4\right)
+\left(\binom{n+1}4-\binom{n}4\right)
=\binom{n+3}4-3\binom{n+2}4+3\binom{n+1}4-\binom{n}4
$$
etc.
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