(6 6 + 14)²ⁿ⁺¹= p

Here, to get the integral part we need to consider only the additon of those terms in which the power of 6 6 is even, as even power of 6 6 will be an even number and this when multiplied by some power of 14 will again give even integer and finally addition of such even terms will also be even.

As far as second part is concerned, it seems that there is problem or error in it for the reason that you have mentioned that p(f), where f is a fractional part and you are asking to prove a fraction part equal to 20²ⁿ⁺¹ which in turn comes out to be an even integer. So there is a discrepancy in the problem. Kindly check the same