Be careful, spoiler! I am going to explain the full proof below, if you want to try it by yourself follow the advice of @nponeccop and try to expand your function calls ;)
Proof
Knowing that:
f . g = \x -> f (g x)
pair :: (a -> b, a -> c) -> a -> (b, c)
pair (f, g) x = (f x, g x)
And that the infix composition operator .
has a lower precedence than function application, you can work out the following:
pair (f, g) . h
= (pair (f, g)) . h -- explicit precedence
= \x -> (pair (f, g)) (h x) -- expanding the composition operator
= \x -> (f (h x), g (h x)) -- expanding 'pair'
= \x -> ((f . h) x, (g . h) x) -- using the composition operator
= \x -> pair (f . h, g . h) x -- back to 'pair'
= pair (f . h, g . h)
Q.E.D if I didn't make a boo boo... Hope this helped!