Finding a more intuitive way to approach a summation of factorials 
I encountered a question in my exam they asked us to solve the following summation.
$$\sum_{r=1}^{10}r!(r^3+6r^2+2r+5)$$
I have found a solution of this online where they express the cubic expression as-
$$(r+1)(r+2)(r+3)-9(r+1)+8$$
JEE Mains 2021, March 18th evening attempt

After proceeding from this I can solve it too.
My question is why this intuition works; we can write the same expression in many ways but only the above manipulation works. How can one figure out that writing the expression as above will work in limited time? Also, is there any other way to tackle this question?
 A: The intuition here is that , in an examination setting , most of questions do not require much tedious work including very large calculations . In questions regarding the sum of factorials ,it's most likely is to be solved by a sort of telescopic method   , which after writing the terms , a lot of terms cancel out , with 2 or 3 terms left only at the end .
I think it's from the JEE main examination , and this question also came  in my exam .
A: In summations, a very convenient trick is to write things like
$$n^2=n(n-1)+n$$
$$n^3=n(n-1)(n-2)+3n(n-1)+n$$
$$n^4=n(n-1)(n-2)(n-3)+6n(n-1)(n-2)+7n(n-1)+n$$
A: The intuition behind this is that
$$n!(n+1)(n+2)\ldots(n+k)=(n+k)!$$
Hence we can transform the polynomial multiplied with the factorial to get a sum of factorials, which may telescope.
Note that the expression $n(n+1)(n+2)(n+3)\ldots (n+k)$ is known as the rising factorial, which is denoted as $n^{(k+1)}$.
To find the expression for $r^3+6r^2+2r+5$ in terms of rising factorials of $r+1$, we can systematically find the coefficient of $(r+1)^{(3)}$ that matches with the coefficient of $r^3$ in $r^3+6r^2+2r+5$ and so on.
We have
$$(r+1)^{(3)}=(r+1)(r+2)(r+3)=r^3+6r^2+11r+6$$
$$(r+1)^{(2)}=(r+1)(r+2)=r^2+3r+2$$
$$(r+1)^{(1)}=r+1$$
$$(r+1)^{(0)}=1$$
The coefficients of these expansions are the stirling numbers of the first kind.
Note that in the equation
$$r^3+6r^2+2r+5=a(r+1)^{(3)}+b(r+1)^{(2)}+c(r+1)^{(1)}+d(r+1)^{(0)}$$
comparing the coefficients of $r^3$ gives $a=1$. We then have
$$r^3+6r^2+2r+5=r^3+6r^2+11r+6+b(r+1)^{(2)}+c(r+1)^{(1)}+d(r+1)^{(0)}$$
$$-9r-1=b(r+1)^{(2)}+c(r+1)^{(1)}+d(r+1)^{(0)}$$
Using similar reasoning it follows that $b=0,c=-9,d-8$.
Another way to solve this would be by finding the expression for any $r^n$ in linear combinations of rising factorials of $(r+1)^{(k)}$. We can use the well known property that
$$r^n=\sum_{k=0}^n \begin{Bmatrix}n\\k\end{Bmatrix}(-1)^{n-k}r^{(k)}$$
It should follow relatively straight forward from there.
A: As @Filthyscrub also told in the comments, it has to do with factorial properties.

 $$(r+1)(r+2)(r+3)\cdot r!=(r+3)!$$
$$-(r+1)\cdot r!=-(r+1)!$$

Since there is one factorial term with a positive sign, and another smaller factorial term with negative sign, when you expand the summation, factorial terms will cancel out. This will just ease the solution. This is called telescoping series.
Other methods are tedious because the summation of factorials is not calculated easily.
In fact, whenever you find any summation of a factorial with some polynomial, you can use such a factorisation. The answer by @ClaudeLeibovici may help in that.
Hope this helps. Ask anything if not clear :)
