Is factorial considered a closed formula? I know this is a tricky question, but I'm curious if factorial is truly considered a closed formula? We all know that:
$$
\prod_{i=1}^n i = n!
$$
I've always considered the LHS to not be a closed formula and the RHS to be a closed formula. But from another viewpoint, is the exclamation point just a notation for the LHS? If this is the case, can we truly consider the RHS to be a closed formula? Moreover, it's just shorthand.
The same goes for a summation. In my head, I've always believed a closed formula did not include any iteration or recursion, even if it's finite. An example of iteration is Gauss' sum of consecutive integers from $1$ to $n$:
$$
\sum_{i=1}^n i = \frac{n \left( n + 1 \right)}{2}
$$
Again, the LHS is not a closed formula, but the RHS is. An example of recursion is the Fibonacci sequence:
$$
F_n = F_{n-1} + F_{n-2} = \frac{1}{\sqrt{5}} \left( \varphi^n - \phi^n \right)
$$
The middle is not a closed formula, but the RHS is. I know there are tons of examples, but these are just a few simple examples off the top of my head.
So, is factorial really a closed formula? If someone derived a formula that included a factorial, can we consider that to be a closed formula? In fact, are binomial coefficients closed formulas, since:
$$
\binom{n}{k} = \frac{n!}{k! \left( n - k \right)!}
$$
Any insight to this will greatly be appreciated. Thank you!
 A: Based on the subject and the person, whether the factorial is considered closed-form or not would differ. For example, in combinatorics, the factorial is used so regularly that it is basically considered closed form since the factorial is a standard operation. However, in subjects like real analysis, the factorial is used sparingly enough to not be considered as a standard operation, and it would not be treated as closed form. It is however used in Taylor series expansions, so it is useful even in real analysis.
Also, the notion of closed-form-ness is not very useful now, since it is meant for ease of approximation and computers can approximate basically anything. In this video, at 14:08, James Grime claims that the approximation was "easier to work out," and at 14:29 "if you wanted to work it out, you'd be better off approximating." Even though that particular case uses very complicated formulas, this might be true in other cases as well.
A: A helpful discussion of this aspect is given in in section 1.2 in Concrete Mathematics by R.L. Graham, D. Knuth and O. Patashnik. We need some formulas which are cited in the text below.
\begin{align*}
T_0&=0\\
T_n&=2T_{n-1}+1\qquad\qquad\quad\ \ \text{for}\  n>0\tag{1.1}\\
\\
T_n&=2^{n}-1\qquad\qquad\qquad\quad\text{for}\  n\geq 0\tag{1.2}\\
\\
L_0&=1\\
L_n&=L_{n-1}+n\qquad\qquad\qquad\text{for}\  n>0\tag{1.4}\\
\\
L_n&=\frac{n(n+1)}{2}+1\qquad\qquad\;\text{for}\  n\geq 0\tag{1.6}\\
\end{align*}

The following is verbatim from section 1.2:

*

*Incidentally we've been talking about closed forms without explicitly saying what we mean. Usually it's pretty clear. Recurrences like (1.1) and (1.4) are not in closed form - they express a quantity in terms of itself; but solutions like (1.2) and (1.6) are. Sums like $1+2+\cdots+n$ are not in closed form - they cheat by using ' $\cdots$'; but expressions like $n(n+1)/2$ are.


*We could give a rough definition like this: An expression for a quantity $f(n)$ is in closed form if we can compute it using at most a  fixed number of well known standard operations, independent of $n$. For example, $2n-1$ and $n(n+1)/2$ are closed forms because they involve only addition, subtraction, multiplication, division, and exponentiation, in explicit ways. The total number of simple closed forms is limited, and there are recurrences that don't have simple closed forms. When such recurrences turn out to be important, because they arise repeatedly, we add new operations to our repertoire; this can greatly extend the range of problems solvable in simple closed form.


*For example, the product of the first $n$  integers, $n!$, has proved to be so important that we now consider it a basic operation. The formula $n!$ is therefore in closed form, although its equivalent $1\cdot 2\cdots n$ is not.

We conclude from these explanations: The right-hand side of the identity
\begin{align*}
\color{blue}{\prod_{i=1}^ni=n!}
\end{align*}
is given in closed form whereas the left-hand side is not in closed form. Whenever we have bound variables as the index $i$ in the product above, we don't have a closed form representation.

Richard P. Stanley indicates this in Enumerative Combinatocis section 1.1 when he tells us:

*

*The most satisfactory form of $f(i)$ is a complete explicit closed formula involving only well-known functions, and free from summation symbols.


For more information on this extract, see here.
