Let

fix

(where

is just the set of all

-degree polynomials of degree

) and let

be the mapping

. This mapping is linear since

and thus the function

where

is linear. But, since

is a basis for

; to prove that

for all

it suffices to check it for

. To do this we merely note that

where we've used the fact that after the

term the terms vanish (that's how we can account for the change of the upper index) and at the last step we employed the Binomial theorem. The conclusion follows by noticing that

is the

degree Taylor polynomial at

, where of course we realize that the remainder term is zero since the

derivative of

is identically zero.