Can there be only one extension to the factorial? Usually, when someone says something like $\left(\frac12\right)!$, they are probably referring to the Gamma function, which extends the factorial to any value of $x$.
The usual definition of the factorial is $x!=1\times2\times3\times\dots x$, but for $x\notin\mathbb{N}$, the Gamma function results in $x!=\int_0^\infty t^xe^{-t}dt$.
However, back a while ago, someone mentioned that there may be more than one way to define the factorial for non-integer arguments, and so, I wished to disprove that statement with some assumptions about the factorial function.

the factorial


*

*is a $C^\infty$ function for $x\in\mathbb{C}$ except at $\mathbb{Z}_{<0}$ because of singularities, which we will see later.

*is a monotone increasing function that is concave up for $x>1$.

*satisfies the relation $x!=x(x-1)!$

*and lastly $1!=1$

From $3$ and $4$, one can define $x!$ for $x\in\mathbb{N}$, and we can see that for negative integer arguments, the factorial is undefined.  We can also see that $0!=1$.
Since we assumed $2$, we should be able to sketch the factorial for $x>1$, using our points found from $3,4$ as guidelines.
At the same time, when sketching the graph, we remember $1$, so there can be no jumps or gaps from one value of $x$ to the next.
Then we reapply $3$, correcting values for $x\in\mathbb{R}$, since all values of $x$ must satisfy this relationship.
Again, because of $1$, we must re-correct our graph, since having $3$ makes the derivative of $x!$ for $x\in\mathbb N$ undefined.
So, because of $1$ and $3$, I realized that there can only be one way to define the factorial for $x\in\mathbb R$.
Is my reasoning correct?  And can there be only one extension to the factorial?

Oh, and here is a 'link' to how I almost differentiated the factorial only with a few assumptions, like that it is even possible to differentiate.
Putting that in mind, it could be possible to define the factorial with Taylor's theorem?
 A: First, for a fixed $c\in\mathbb{C}$, let $$F_c(z):=\Gamma(z+1)\cdot\big(1+c\,\sin(2\pi z)\big)$$ for all $z\in\mathbb{C}\setminus \mathbb{Z}_{<0}$, which defines an analytic function $F_c:\mathbb{C}\setminus\mathbb{Z}_{<0}\to\mathbb{C}$ such that $$F_c(z)=z\cdot F_c(z-1)$$
for all $z\in\mathbb{C}\setminus\mathbb{Z}_{\leq 0}$ and that $F_c(0)=F_c(1)=1$ (whence $F_c(n)=n!$ for every $n\in\mathbb{Z}_{\geq 0}$).  Excluding the essential singularity at $\infty$, the negative integers are the only singularities of $F_c$, which are simple poles.   
Here are some results I checked with Mathematica.  If $c$ is a positive real number less than $0.022752$, then $F_c'(z)>0$ for all $z>1$ and $F_c''(z)>0$ for all $z>-1$, making $F_c$ monotonically increasing on $(1,\infty)$ and convex on $(-1,\infty)$.  It also appears that, with $0<c<0.022752$, $F_c$ is convex on $(-2n,-2n+1)$ and concave on $(-2n-1,-2n)$ for every $n=1,2,\ldots$.  (I have checked this with various values of $c$ and with $n\leq 30$.)  It would be great if someone can find an actual proof.  Hence, it seems to me that the conditions 1-4 do not give a unique factorial function.
A: The Bohr-Mollerup Theorem states that the Gamma function is the only log-convex function which satisfies $\Gamma(1)=1$ and $x\Gamma(x)=\Gamma(x+1)$. There are a continuum of functions that are not log-convex that satisfy the other two constraints.
This answer details some of the important points of this.
