Can the factorial function be written as a sum? I know of the sum of the natural logarithms of the factors of n! , but would like to know if any others exist.
 A: Sure, for $n\ge 1$ we have $$n!=\fbox{$(n-1)\times (n-1)!$} + \fbox{$(n-1)!$}$$
A: This one is pretty important:
$$n! = \sum_{\sigma\in S_n} 1$$
Edit: As Arkamis explains, $S_n$ is the symmetric group on $n$ letters. Each $\sigma\in S_n$ is a permutation on the set $[1,2,\ldots,n]$. Since $S_n$ is a finite set, we may sum a function over it, and the sum of the constant function $f(\sigma)=1$ is just the size of the set, which is $|S_n| = n!$.
Arguably, summing a constant function is cheating. Here's one way to raise the stakes. Let $B_n$ be the set of $n\times n$ integer matrices $A$ such that every sum of a subset of entries from $A$ is in $[0,n]$. Then:
$$n!=\sum_{A\in B_n}|\det A|$$
This is the same identity in a more interesting disguise. Every $n\times n$ permutation matrix $A$ is a member of $B_n$, and $\det A = \pm 1$. On the other hand, if $A\in B_n$ is not a permutation matrix, then you can prove that $\det A = 0$.
A: If you really want, we can write $n!$ using this infinite series
A: $$n!=\prod_{p<n}p^{\sum_{k=1}^\infty\lfloor{n\over p^k}\rfloor}$$
A: Another combinatorially important sum:
$\displaystyle n! = \sum_{k=0}^n\bigl[\begin{smallmatrix}n\\k \end{smallmatrix}\bigr]$, where $\displaystyle\bigl[\begin{smallmatrix}n\\k \end{smallmatrix}\bigr]$ is the (unsigned) Stirling symbol of the first kind, which can be defined (for instance) by the relation $x(x+1)\ldots(x+n-1) = \sum_{k=0}^n\bigl[\begin{smallmatrix}n\\k \end{smallmatrix}\bigr] x^k$. (And of course, by setting $x=1$ in this relation we get the initial sum.)  The combinatorial significance is that $\bigl[\begin{smallmatrix}n\\k \end{smallmatrix}\bigr]$ counts the number of permutations of the numbers $1..n$ with $k$ distinct cycles; the relation $\displaystyle n! = \sum_{k=0}^n\bigl[\begin{smallmatrix}n\\k \end{smallmatrix}\bigr]$ thus says that the total number of permutations is just the sum over all $k$ of the number of permutations into $k$ cycles.
A: I have found the following formula:
$$n!=\sum_{k=2}^{n+1}(-1)^{n+1-k}\binom{n+1}{k}\sum_{i=1}^{k-1}i^{n},\ n\in N.$$
A: I know this question is old, but I found it googling for summation definitions of the factorial function, and thought a good answer would be the following formula, which I discovered several years ago and I haven't seen anywhere else:
$n!=\displaystyle\sum_{k=1}^n \binom{n}{k} (-1)^{n+k} k^n$
I never took the time to find a proof, but it certainly works. It would be interesting if anyone with decent math skills (that is, better than me) could shed some light on it.
A: Lots of good answers, but the answer is quite easily, "yes."
Integer multiplication is repeated addition.
$n!$ is $n$ groups of $(n-1)!$ objects, so $n! = \sum_{k=1}^n (n-1)!$.
Then, $(n-1)!$ is $n-1$ groups of $(n-2)!$ objects, so $n! = \sum_{k_1=1}^n \sum_{k_2 = 1}^{n-1} (n-2)!$, and so on.

Example:
$$4! = \sum_{k_1 = 1}^4 3! = 3!+3!+3!+3! = 4\cdot 3!.$$
$$\begin{align*}
4! &= \sum_{k_1=1}^4 \sum_{k_2=1}^3 2! \\
&= \sum_{k_1=1}^4 2!+2!+2! \\
&= 2!+2!+2! + 2!+2!+2! + 2!+2!+2! + 2!+2!+2! \\
&= 12\cdot 2! \\
&=4\cdot 3\cdot 2!. \end{align*}$$
A: $$n! = 1 + \sum_{k=1}^n (k-1) (k-1)!$$
A: $n! = e^{ \sum_{k = 1}^n \ln (k)}$
A: $$\frac{(n+2)!}{2} =  (n+1)! + \sum_{i=1}^n i\,n!$$
A: Here's a good reference document from a few years back: Factorials as sums by Roberto Anglani and Margherita Barile.

In this paper we give an additive representation of the factorial, which can be proven by a simple quick analytical argument. We also present some generalizations, which are linked, on the one hand to an arithmetical theorem proven by Euler (decomposition of primes as the sum of two squares), and, on the other hand, to modern combinatorics (Stirling numbers).

A: $$(v+1)! = 1 + \sum_{k=1}^v (v-k-1)! (v-k)^2$$
