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 $n$

$\sum_{k=0}^n\dfrac{(-1)^k (-k)^n}{k!\,(n - k)!}=1$

Pi function

Finite product of complex numbers