Is the gamma function $\Gamma(n+1)$ the only continuous function and defined derivative with the same recursive definition as of $n!$ for $n>-1$?

Question

Is the gamma function $\Gamma(n+1)$ the only continuous function and defined derivative with the same recursive definition as of $n!$ for $n>-1$ ? (When using real numbers.)

The recursive definition of $n!$ I'm referring to:

$0!=1$
$n!=n\cdot(n-1)!\quad$ for $\quad n\in\mathbb Z^+$

Clarification: let $y$ be a function with the restrictions:

$y(0)=1$
$\forall x\in\mathbb R^+:y(x)=xy(x-1)$

Find a continuous function $y(x)$ in the intervall $x>-1$ with a defined $y'(x)$ in the intervall $x>-1$. Is $y(x)$ always the function $\Gamma(x+1)$ ?



Notice: I edited $\Gamma(x-1)$ to $\Gamma(x+1)$.

Answer 1

Unless the function is logarithmically convex, it won't be unique (the Gamma function). See the Bohr-Mollerup theorem.

Answer 2

Also for a function/sequence $f(n) = a_n$, consider $f(n)\cos(2\pi\,n) = a_n$ for integer values of $n$.