Is there a rule to the terms of a falling factorial? $\require{cancel}$I discovered that $n!=\xcancel{(n)_{n-1}}n^{\underline{n-1}}=n(n-1)(n-2)\cdots(3)(2)$. I have expanded a few examples:
$$2!=\xcancel{(2)_1}2^{\underline{1}}=2\\
3!=\xcancel{(3)_2}3^{\underline{2}}=3(3-1)=3^2-3=9-3=6\\
4!=\xcancel{(4)_3}4^{\underline{3}}=4(4-1)(4-2)=4^3-3\cdot4^2+2\cdot4=64-48+8=24\\
5!=\xcancel{(5)_4}5^{\underline{4}}=5(5-1)(5-2)(5-3)=5^4-6\cdot5^3+11\cdot5^2-6\cdot5\\=625-750+275-30=120$$
My question is what rule is there to determine the proper coefficients for the expansion of the falling factorial? For example, let's say we have coefficients $a, b, c, d, e$ such that
$$\xcancel{(x)_8}x^{\underline{8}}=x^8-\binom{8}{2}x^7+ax^6-bx^5+cx^4-dx^3+ex^2-7!x$$
What rule enables me to choose those coefficients without multiplying the factors together?
Incidentally, the full polynomial is
$$\xcancel{(x)_8}x^{\underline{8}}=x^8-28x^7+322x^6-1960x^5+6769x^4-13132x^3+13068x^2-5040x$$
Edit: Since there was initially some confusion as to the notation's meaning, I have canceled the previous notation and added other notation that has (I hope) only one use. I hope this clears things up. I definitely meant Falling Factorial, but as it has been brought to my attention that the symbol I chose may have misled some people, I thought this change might be necessary.

The present discussion is polluted by the confusion of unfortunate symbols for falling and rising factorials as well as the Pochhammer symbol. This has been well underlined in the comments. In order to make it clear to everybody, an excerpt from WolframMathWorld is attached below.
So, without using the Pochhammer symbol, the formula is :
$$ x(x+1)(x+2)…(x+n-1) = s(n,n)x^n-s(n,n-1)x^{n-1}+…+(-1)^{n-1}s(n,1)x  $$
$$ x(x-1)(x-2)…(x-n+1) = s(n,n)x^n+s(n,n-1)x^{n-1}+…+s(n,1)x  $$
where $s(n,k)$ is the Stirling number of the first kind.
Using the Stirling’s symbol $s(n,k)$ is the simplest way to represent analytically the coefficients of the polynomial. Of course, if the goal is to simplify the numerical computation, using the Stirling numbers  is not an advantage in the practical viewpoint.

