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