Can anyone explain me this one?

The question is how to show that the above two products are equal to each other.

I don't know if this is the best way, but this is how I would do it:

$\large {\prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k=1+m}^{k=n+m}a_k$     because  $\prod\limits_{k=1+m}^{n+m}a_k$   means the same as  $\prod\limits_{k=1+m}^{k=n+m}a_k$

$\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k-m=1}^{k-m=n}a_k$      by subtracting  m  from both sides of each equation

$\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k-m=1}^{k-m=n}a_{(k-m+m)}$         since  k - m + m  =  k

$\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{{\color{RubineRed}k-m}=1}^{{\color{RubineRed}k-m}=n}a_{({\color{RubineRed}k-m}+m)}$

$\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k=1}^{k=n}a_{(k+m)}$       because we can replace the pink text with whatever we want

$\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k=1}^{n}a_{(k+m)}$

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