# Proof the equality $\prod_{r=1}^{mn}\left(x+mn -r\right)=\prod_{k=1}^{m}\prod_{l=1}^{n}\left(x+mn-(1+ml-k) \right)$

I came across a proof of Gauss multiplication formula for the Gamma function which relies on the following indentity (without a proof)

$$\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=m^{mn}\prod_{k=1}^{m}\frac{\Gamma \left(\frac{x+k-1}{m}+n \right)}{\Gamma \left(\frac{x+k-1}{m} \right)} \, \tag{1}$$

I am trying to prove it.

I started expanding the left hand side first. From the recurrence relation of the Gamma function we have that

$$\Gamma \left(x+mn \right)= \left(x+mn -1\right)\Gamma \left(x+mn-1 \right)$$

$$\Gamma \left(x+mn \right)= \left(x+mn -1\right) \left(x+mn -2\right)\Gamma \left(x+mn-2 \right)$$

$$\cdots$$

$$\Gamma \left(x+mn \right)= \left(x+mn -1\right) \left(x+mn -2\right) \cdots \left(x+mn -mn\right)\Gamma \left(x+mn-mn \right)$$

Therefore we can rewrite the L.H.S of $$(1)$$ as

$$\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=\left(x+mn -1\right) \left(x+mn -2\right) \cdots \left(x\right)$$

$$\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=\prod_{r=1}^{mn}\left(x+mn -r\right) \, \tag{2}$$

Similarly, for the R.H.S. of $$(1)$$ we obtain

$$\frac{\Gamma \left(\frac{x+k-1}{m}+n \right)}{\Gamma \left(\frac{x+k-1}{m} \right)}=\left(\frac{x+k-1}{m}+n-1 \right)\left(\frac{x+k-1}{m}+n-2 \right) \cdots \left(\frac{x+k-1}{m}+n-n \right)$$

$$\frac{\Gamma \left(\frac{x+k-1}{m}+n \right)}{\Gamma \left(\frac{x+k-1}{m} \right)}=\prod_{l=1}^{n}\left(\frac{x+k-1}{m}+n-l \right)$$

$$\frac{\Gamma \left(\frac{x+k-1}{m}+n \right)}{\Gamma \left(\frac{x+k-1}{m} \right)}=\prod_{l=1}^{n}\frac{1}{m}\left(x+k-1+mn-ml \right) \, \tag{3}$$

Plugging $$(2)$$ and $$(3)$$ in $$(1)$$ we obtain the following equality.

$$\prod_{r=1}^{mn}\left(x+mn -r\right)=\prod_{k=1}^{m}\prod_{l=1}^{n}\left(x+k-1+mn-ml \right) \, \tag{4}$$

$$\prod_{r=1}^{mn}\left(x+mn -r\right)=\prod_{k=1}^{m}\prod_{l=1}^{n}\left(x+mn-(1+ml-k) \right)$$

suposse $$m=2 \,\, \text{and}\,\,n=2$$, the right hand side becomes

$$\prod_{k=1}^{2}\prod_{l=1}^{2}\left(x+k-1+4-2l \right)=\prod_{k=1}^{2}\left(x+k-1+4-2 \right)\left(x+k-1+4-2\times2 \right)$$

$$=\prod_{k=1}^{2}\left(x+k+1 \right)\left(x+k-1 \right)$$

$$=\left(x+1+1 \right) \cdot\left(x+1-1 \right)\cdot\left(x+2+1 \right)\cdot\left(x+2-1 \right)$$

$$=\left(x+2 \right)\cdot x \cdot \left(x+3 \right) \cdot\left(x+1 \right)$$

$$= x \cdot\left(x+1 \right)\cdot \left(x+2 \right)\cdot \left(x+3 \right) \, \tag{5}$$

And the left hand side becomes

$$\prod_{r=1}^{4}\left(x+4 -r\right)=\left(x+4 -1\right)\left(x+4 -2\right)\left(x+4 -3\right)\left(x+4 -4\right)$$

$$= x \cdot\left(x+1 \right)\cdot \left(x+2 \right)\cdot \left(x+3 \right)$$

Which equals exactly $$(5)$$. So intuitively I see that the equality $$(1)$$ holds. My question is: How can I go from this heuristic intuitive proof to a formal proof, may be proved by induction?

Just notice that any natural number $$\le nm$$ can be uniquely written in the form $$ml - k’$$, where $$1\le l \le n$$, $$0 \le k’ < m$$ are integers (this is just a division by $$m$$ with remainder). Changing $$k’$$ on $$k - 1$$, we can see, that each factor of the left product in (4) appears exactly one time in the right product. But they both have $$mn$$ factors. Hence the products are equal.

Ok, after a long arduous work I got it. Consider $$(1)$$ above

$$\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=m^{mn}\prod_{k=1}^{m}\frac{\Gamma \left(\frac{x+k-1}{m}+n \right)}{\Gamma \left(\frac{x+k-1}{m} \right)} \, \tag{1}$$

Note that both sides of (1), the ration between Gamma functions, can be rewritten as Pocchammer symbols. By the recurrence equation of the Gamma function, namely

$$\Gamma \left(x+1 \right)=x\Gamma \left(x+1 \right)$$

we may write

$$\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=\frac{\left(x+mn-1 \right)\Gamma \left(x+mn-1 \right)}{\Gamma \left(x \right)}$$

$$\cdots$$

$$=\frac{\left(x+mn-1 \right)\left(x+mn-2 \right) \cdots x\Gamma \left(x \right)}{\Gamma \left(x \right)}$$

$$\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=x (x+1) \cdots\left(x+mn-2 \right)\left(x+mn-1 \right)$$

$$\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=\left( x\right)_{mn}\, \tag{2}$$

enforcing $$x \longrightarrow \frac{x+k-1}{m} \, \, \text{and}\,\, mn \longrightarrow n$$ in $$(2)$$ above we get

$$\frac{\Gamma \left(\frac{x+k-1}{m}+n \right)}{\Gamma \left(\frac{x+k-1}{m} \right)}=\left(\frac{x+k-1}{m} \right)_{n}\, \tag{3}$$

plugging $$(2)$$ and $$(3)$$ in $$(1)$$ we get

$$\left( x\right)_{mn}=m^{mn}\prod_{k=1}^{m}\left(\frac{x+k-1}{m} \right)_{n}\tag{4}$$

Now, lets play a little with Pocchammer symbol to see whether we can draw something from it. By definition we have

$$\left( x\right)_{m}=\underbrace{x(x+1)(x+2) \cdots (x+m-2)(x+m-1)}_{\text{n terms}}$$

We have one group of m terms each

$$\left( x\right)_{2m}=\color{red} {x(x+1)(x+2) \cdots (x+m-2)(x+m-1)}\color{blue}{(x+m-1+1)(x+m-1+2)\cdots(x+m-1+m)}$$

$$\left( x\right)_{2m}=\underbrace{\underbrace{\color{red} {x(x+1)(x+2) \cdots (x+m-2)(x+m-1)}}_{\text{n terms}}\underbrace{\color{blue}{(x+m)(x+m+1)\cdots(x+m+m-1)}}_{\text{n terms}}}_{\text{2n terms}}$$

we have 2 groups of m terms each

$$\left( x\right)_{3m}=\underbrace{\underbrace{\color{red} {x(x+1)(x+2) \cdots (x+m-2)(x+m-1)}}_{\text{n terms}}\underbrace{\color{blue}{(x+m)(x+m+1)\cdots(x+m+m-1)}}_{\text{n terms}}\color{green} {\underbrace{(x+2m)(x+2m+1) \cdots(x+2m+m-1)}_{\text{n terms}}}}_{\text{3n terms}}$$

We have 3 groups of m terms each

Generalizing we may write

$$\left( x\right)_{nm}=\color{red} {x(x+1)(x+2) \cdots (x+m-2)(x+m-1)} \color{black}{\cdots} \color{blue}{(x+(n-1)m)(x+(n-1)m+1)\cdots(x+(n-1)m+m-1)}$$

$$\left( x\right)_{nm}=\color{red} {x(x+1)(x+2) \cdots (x+m-2)(x+m-1)} \color{black}{\cdots} \color{blue}{(x+(n-1)m)(x+(n-1)m+1)\cdots(x+nm-1)}$$

We obtain n groups of m terms each.

Now, instead o writing each group horizontally, lets write each group in a different line getting sort of a matrix look. For $$(x)_{2m}$$

\begin{aligned} &\begin{array}{llcclll} (x)_{2m}=&x & (x+1) & (x+2) & \cdots & (x+m-1) \\ &\left(x+m\right) & \left(x+m+1\right) & \left(x+m+2\right) & \ldots & \left(x+m+m-1 \right) \\ \end{array}\\ \end{aligned}

\begin{aligned} &\begin{array}{llcclll} (x)_{2m}=&m^m& \frac{x}{m} & (\frac{x+1}{m}) & (\frac{x+2}{m}) & \cdots & (\frac{x+m-1)}{m} \\ &m^m&\left(\frac{x}{m}+1\right) & \left(\frac{x+1}{m}+1\right) & \left(\frac{x+2}{m}+1\right) & \ldots & \left(\frac{x+m-1}{m}+1 \right) \end{array}\\ \end{aligned}

Now multiply vertically along each column to obtain

$$(x)_{2m}=m^{2m} \cdot \left(\frac{x}{m}\right)_{2} \cdot\left(\frac{x+1}{m}\right)_{2} \cdot \left(\frac{x+2}{m}\right)_{2} \cdots \left(\frac{x+m-1}{m}\right)_{2}$$

And we may write

$$(x)_{2m}=m^{2m}\prod_{k=0}^{m-1}\left(\frac{x+k}{m} \right)_{2}$$

Similarly

\begin{aligned} &\begin{array}{llcclll} (x)_{3m}=&x & (x+1) & (x+2) & \cdots & (x+m-1) \\ &\left(x+m\right) & \left(x+m+1\right) & \left(x+m+2\right) & \ldots & \left(x+m+m-1 \right) \\ &\left(x+2m\right) & \left(x+2m+1\right) & \left(x+2m+2\right) & \ldots & \left(x+2m+m-1 \right) \end{array}\\ \end{aligned}

\begin{aligned} &\begin{array}{llcclll} (x)_{3m}=&m^m& \frac{x}{m} & (\frac{x+1}{m}) & (\frac{x+2}{m}) & \cdots & (\frac{x+m-1)}{m} \\ &m^m&\left(\frac{x}{m}+1\right) & \left(\frac{x+1}{m}+1\right) & \left(\frac{x+2}{m}+1\right) & \ldots & \left(\frac{x+m-1}{m}+1 \right) \\ &m^m&\left(\frac{x}{m}+2\right) & \left(\frac{x+1}{m}+2\right) & \left(\frac{x+2}{m}+2\right) & \ldots & \left(\frac{x+m-1}{m}+2 \right) \\ (x)_{3m}=&m^{3m}& \left(\frac{x}{m}\right)_{3}&\left(\frac{x+1}{m}\right)_{3}&\left(\frac{x+2}{m}\right)_{3}& \cdots &\left(\frac{x+m-1}{m}\right)_{3} \end{array}\\ \end{aligned}

And we may write

$$(x)_{3m}=m^{3m}\prod_{k=0}^{m-1}\left(\frac{x+k}{m} \right)_{3}$$

Generalizing, we may write

$$(x)_{nm}=A_{1} \cdot A_{2} \cdots A_{n}$$

where

$$\begin{array}{cccccc} A_{1}= & x & (x+1) & (x+2) & \ldots & (x+m-1) \\ A_{2}= & (x+m) & (x+m+1) & (x+m+2) & \ldots & (x+m+ m-1) \\ A_{3}= & (x+2 m) & (x+2 m+1) & (x+2 m+2) & \ldots & (x+2m+ m-1) \\ \vdots & \vdots & \vdots & \ldots & \ldots & \vdots \\ A_{n}= & (x+(n-1) m) & (x+(n-1) m+1) & (x+(n-1) m+2) & \ldots & (x+(n-1)m+ m-1) \end{array}$$

Factoring out m in each line

\begin{aligned} &\begin{array}{llcclll} A_{1}= & m^{m} & \frac{x}{m} & \left(\frac{x+1}{m}\right) & \left(\frac{x+2}{m}\right) & \ldots & \left(\frac{x+m-1}{m}\right) \\ A_{2}= & m^{m} & \left(\frac{x}{m}+1\right) & \left(\frac{x+1}{m}+1\right) & \left(\frac{x+2}{m}+1\right) & \ldots & \left(\frac{x+m-1}{m}+1\right) \\ A_{3}= & m^{m} & \left(\frac{x}{m}+2\right) & \left(\frac{x+1}{m}+2\right) & \left(\frac{x+2}{m}+2\right) & \ldots & \left(\frac{x+ m-1}{m}+2\right)\\ \vdots & \vdots & \vdots & \ldots & \ldots & \vdots\\ A_{n}=&m^{m} &\left(\frac{x}{m}+n-1\right)& \left(\frac{x+1}{m}+n-1\right) &\left(\frac{x+2}{m}+n-1\right)& \ldots &\left(\frac{x+m -1}{m}+n-1\right)\\ \end{array}\\ \end{aligned}

Which analogously as the above cases, we may then write

$$(x)_{nm}=m^{nm}\left(\frac{x}{m} \right)_{n}\left(\frac{x+1}{m} \right)_{n} \cdots \left(\frac{x+m-1}{m} \right)_{n}$$

and finally

$$(x)_{nm}=m^{nm}\prod_{k=0}^{m-1}\left(\frac{x+k}{m} \right)_{n}$$

or shifting the index we get exactly $$(4)$$

$$(x)_{nm}=m^{nm}\prod_{k=1}^{m}\left(\frac{x+k-1}{m} \right)_{n}$$