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.
