Okay I'm not sure if this method is rigorous, seems rather primitive without some formal definitions, but anyways:
Writing the first few terms of the product out you get:
(7)}{(3)(3)}\right) \left(\frac{(2)(13)}{(4)(7)}\right) \left(\frac{(3)(21)}{(5)(13)}\right) \left(\frac{(4)(31)}{(6)(21)}\right) \left(\frac{(5)(43)}{(7)(31)}\right) \left(\frac{(6)(57)}{(8)(43)}\right) ...)
A pattern can be seen with the canceling:
Say we have the fractions in the form of
Then the cancelling looks like this: (Ignore the first two fractions of the product series, consider from
(21)}{(5)(13)})
onwards...)
To cancel out the top right number, it will correspond with the bottom right number of the following fraction:
(b)}{()()}\right)\left(\frac{()()}{()(b)}\right))

'

' will always cancel.
To cancel the top left number, it will always correspond with the bottom left number of 2 fractions BEFORE it.
()}{(a)()}\right)...\left(\frac{(a)()}{()()}\right))
where

represents 1 other fraction in between.

'

' will always cancel.
Therefore since the top 2 numbers of any fraction will cancel out the bottom 2 numbers of another fraction, then ALL fractions in the product series to infinity will cancel.
Now analyzing the first two fractions we see some terms which can not be 'canceled'
(7)}{(3)(3)}\right) \left(\frac{(2)(13)}{(4)(7)}\right))
The 1 can be omitted as it does nothing.
Using the "pattern" we formed earlier, both the top right numbers of both fractions will cancel. So we are left with:
(3)}\right) \left(\frac{(2)}{(4)}\right))
Now the top left number, ie the 2 in the second fraction, can not cancel as there is only 1 fraction preceding it, since we realised we need TWO fractions preceding it to cancel it.
Now both the 3 and 4 in the respective fractions can be cancelled since they are the bottom left of the each fraction which means there must be another 2 fractions FOLLOWING it which can cancel them out. Thus 3 and 4 are gone.
Thus we are only left with
}{(3)}\right) \left(\frac{(2)}{(1)}\right))
 = \frac{2}{3})
Is there any other methods with this question? The canceling is not bad but is there any more rigorous methods of doing it?
Thanks!