Proof that $n+k+3$ divides $n(n+1)(n+2)(n+3) - k(k+1)(k+2)(k+3)$. I'm looking for proof that
$$
(n+k+3) \mid n(n+1)(n+2)(n+3) - k(k+1)(k+2)(k+3)\\
n,k \in \mathbb N^*, n>k
$$
I tried using induction, but i'm not sure how it would work with 2 parameters.

Since
$$n+3=n+k+3-k\qquad \text{and}\qquad k+3=n+k+3-n$$
we have
\begin{align}
n(n+1)(n+2)(n+3)&=n(n+1)(n+2)(n+k+3)-n(n+1)(n+2)k\\
k(k+1)(k+2)(k+3)&=k(k+1)(k+2)(n+k+3)-k(k+1)(k+2)n
\end{align}
Then it will be sufficient to show that $n+k+3$ divides $k(k+1)(k+2)n-n(n+1)(n+2)k$
But
\begin{align}
k(k+1)(k+2)n-n(n+1)(n+2)k&=nk(k^2+3k-n^2-3n)\\
&=nk\left[-(n-k)(n+k)-3(n-k)\right]\\
&=-nk(n-k)(n+k+3)
\end{align}
Which completes the proof.
