Equivalence of Definitions of Gamma Function

Theorem
The following definitions of the Gamma function are equivalent:

Weierstrass Form equivalent to Euler Form
First it is shown that the Weierstrass form is equivalent to the Euler form.

Combining the limits:

But:
 * $(1): \quad m = \dfrac {m!} {\left({m - 1}\right)!} = \dfrac 2 1 \cdot \dfrac 3 2 \cdots \dfrac {x+1} x \cdots \dfrac m {m - 1}$

Each term in $(1)$ is just $\dfrac{x+1} x = 1 + \dfrac 1 x$, so:
 * $\displaystyle m = \prod_{n \mathop = 1}^{m-1} \left({1 + \frac 1 n}\right)$

Thus the expression for $\dfrac 1 {\Gamma \left({z}\right)}$ becomes:

Hence:
 * $\displaystyle \Gamma \left({z}\right) = \frac 1 z \prod_{n \mathop = 1}^{\infty} \left({1 + \frac 1 n}\right)^z \left({1 + \frac z n}\right)^{-1}$

which is the Euler form of the Gamma function.

Integral Form equivalent to Euler Form
It is taken for granted that the Gamma function increases monotonically on $\R_{\ge 1}$.

We begin with an inequality that can easily be verified using cross multiplication.

Let $x$ be a real number between $0$ and $1$.

Let $n$ is a positive integer.

Then:
 * $\displaystyle \frac {\log \Gamma \left({n - 1}\right) - \log \Gamma \left({n}\right)} {\left({n - 1}\right) - n} \le \frac {\log \Gamma \left({x + n}\right) - \log \Gamma \left({n}\right)} {\left({x + n}\right) - n} \le \frac {\log \Gamma \left({n + 1}\right) - \log \Gamma \left({n}\right)}{\left({n + 1}\right) - n}$

Since n is a positive integer, we can make use of the identity:
 * $\Gamma \left({n}\right) = \left({n - 1}\right)!$

Simplifying, we get:
 * $\log \left({n - 1}\right) \le \dfrac {\log \Gamma \left({x + n}\right) - \log \left({\left({n - 1}\right)!}\right)} x \le \log \left({n}\right)$

We now make use of the identity:
 * $\displaystyle \Gamma \left({x + n}\right) = \prod_{k \mathop = 1}^n \left({x + n - k}\right) \Gamma \left({x}\right)$

along with the fact that the Gamma Function is Log-Convex, to simplify the inequality:


 * $\displaystyle \left({n - 1}\right)^x \left({n - 1}\right)! \prod_{k \mathop = 1}^n \left({x + n - k}\right)^{-1} \le \Gamma \left({x}\right) \le n^x \left({n - 1}\right)!\prod_{k \mathop = 1}^n \left({x + n - k}\right)^{-1}$

Taking the limit as $n$ goes to infinity and using the Squeeze Theorem:


 * $\displaystyle \Gamma \left({x}\right) = \lim_{n \mathop \to \infty} n^x n! \prod_{k \mathop = 0}^n \left({x + n - k}\right)^{-1}$

which is another representation of Euler's form.

This proves equivalence for $x$ between $0$ and $1$.

The result follows from the Gamma Difference Equation.