Why is there no explicit formula for the factorial?

Question

I am somewhat new to summations and products, but I know that the sum of the first positive n integers is given by:

$$\sum_{k=1}^n k = \frac{n(n+1)}{2} = \frac{n^2+n}{2}$$

However, I know that no such formula is known for the most simple product (in my opinion) - the factorial:

$$n!=\prod_{k=1}^n k = \prod_{k=2}^n k$$

I don't know if that is how products work, but I would really like to know!

So my question is why is there no explicit formula (in terms of n, other than n(n-1)...2*1) for the product of the first n integers? Is there a proof that says that one cannot exist or is it that one has not been discovered?

By explicit formula I mean a non-functional equation that does not require n amount of steps to calculate - just like the summation formula does not require n additions.

Answer 1

Since nobody seems to be willing to answer, let me try to summarize the hints from the comments.

First, we can write factorial as $$n!=\int_0^{\infty}x^n e^{-x}dx.$$ One of the consequences of this formula is Stirling's approximation: as $n\rightarrow\infty$, $$n!\sim \left(\frac{n}{e}\right)^n\sqrt{2\pi n}.\tag{1}$$ At first sight, if there was an explicit algebraic formula for $n!$ in terms of $n$, in such expression there would be no place for $e$ or $\pi$; since they are both present in the asymptotics, one is tempted to conclude that no such algebraic formula can exist. Actually this is not quite true: as was pointed to me by Michalis, one can e.g. obtain $e$ as the asymptotics of $\left(1+\frac{1}{n}\right)^n$.

To show non-existence rigorously, you should first rigorously define what do you mean by "explicit formula" $n!=f(n)$, i.e. to define the class of "acceptable" functions $f$. For example, rational $f$ are immediately ruled out by the above asymptotics.

Answer 2

(The following is a joke.)

Put $a_n:={1\over 6}n(n+1)(n+2)$ and $b_n:={1\over2}n(n+1)$, and define the magic constant $$\xi:=\sum_{n=1}^\infty {n!\over 2^{a_n}}\ =\ 0.630882266676063396815526621896\ldots\quad .$$ Then $$n!=\bigl\lfloor\> 2^{a_n}\xi\>\bigr\rfloor\quad \bigl({\rm mod}\ \ 2^{b_n}\bigr)\qquad(n\geq1)\ .$$ Try it out!