| definiendum | $!: \mathbb N\to \mathbb N$ |
| definiendum | $n\mapsto n!:=\prod_{k=1}^n\ k $ |
Thinking of $n!=\left.\frac{{\mathrm d}^n}{{\mathrm d}x^n}\right|_{x=0}x^n$ and Fermat theory, I though there must be an expression for $n!$ which is more algebraic and indeed I found
Table[Sum[(-1)^k*Binomial[n, k] (-k)^n, {k, 1, n}], {n, 1, 8}]
e.g.
0! = + 0^0 1! = - 0^1 + 1*1^1, 2! = + 0^2 - 2*1^2 + 2^2, 3! = - 0^3 + 3*1^3 - 3*2^3 + 3^3, 4! = + 0^4 - 4*1^4 + 6*2^4 - 4*3^4 + 4^4, 5! = - 0^5 + 5*1^5 - 10*2^5 + 10*3^5 - 5*4^5 + 5^5, 6! = + 0^6 - 6*1^6 + 15*2^6 - 20*3^6 + 15*4^6 - 6*5^6 + 6^6, 7! = - 0^7 + 7*1^7 - 21*2^7 + 35*3^7 - 35*4^7 + 21*5^7 - 7*6^7 + 7^7, 8! = + 0^8 - 8*1^8 + 28*2^8 - 56*3^8 + 70*4^8 - 56*5^8 + 28*6^8 - 8*7^8 + 8^8
The binomial coefficients use the factorial of course, so there's not real computational benefit.
The theorem underlying here is that, for all $n$
$\sum_{k=0}^n\dfrac{(-1)^k (-k)^n}{k!\,(n - k)!}=1$