How can I derive what is $1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$ ??

In general, we have

$$\sum_{k=1}^nk(k+1)(k+2)\dots(k+p)=\frac{n(n+1)(n+2)\dots(n+p+1)}{p+2}$$

Prove by induction,

$$\sum_{k=1}^{n+1}k(k+1)(k+2)\dots(k+p)\\=(n+1)(n+2)\dots(n+p+1)+\sum_{k=1}^nk(k+1)(k+2)\dots(k+p)\\=(n+1)(n+2)\dots(n+p+1)+\frac{n(n+1)(n+2)\dots(n+p+1)}{p+2}\\=\frac{\color{#034da3}{(p+2)}\color{#ee8844}{(n+1)(n+2)\dots(n+p+1)}+\color{#034da3}n\color{#ee8844}{(n+1)(n+2)\dots(n+p+1)}}{p+2}\\=\frac{\color{#ee8844}{(n+1)(n+2)\dots(n+p+1)}\color{#034da3}{(n+p+2)}}{p+2}$$

And easy enough to check for $n=1$ to see it is true.

$$1\times2\times3\times\ldots\times(1+p)=\frac{1\times2\times3\times\ldots\times(1+p)\times\require{cancel}\cancel{(2+p)}}{\cancel{p+2}}$$

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This is slightly off, since my sum ends at $n(n+1)(n+2)(n+3)$, while you end off at $(n-3)(n-2)(n-1)n$. To readjust, have $n=p-3$ in my sum and it will become yours.

Can you prove that $(n-3)(n-2)(n-1)(n) = \frac{(n-3)(n-2)(n-1)(n)(n+1)}{5} - \frac{(n-4)(n-3)(n-2)(n-1)(n)}{5}?$

From there you can substitute different values of $n$ and derive a formula for your expression, and the final summation will be given by

$$ \frac{(n-3)(n-2)(n-1)(n)(n+1)}{5}$$

Following up from Yves Daoust's answer, assume that our sum is a polynomial. Clearly it must be a polynomial of degree $5$ with $1/5$ coefficient because

$$P(n)=\sum_{k=1}^nk(k+1)(k+2)(k+3)\sim\int_1^nx^4+\mathcal O(x^3)dx=\frac15x^5+\mathcal O(x^4)$$

Notice that if we have

$$P(1)=24\qquad P(n)=n(n+1)(n+2)(n+3)+P(n-1)$$

Then

$$P(-4)=P(-3)=P(-2)=P(-1)=P(0)=0$$

Meaning, by the fundamental theorem of algebra,

$$P(n)=\frac{n(n+1)(n+2)(n+3)(n+4)}5$$