A Question on Polynomials, part 2

Let $f(x)= 2\left({x-1 \choose 0} +{x-1 \choose 1}+\ldots +{x-1 \choose 100}\right)$

We see that this the degree of $f$ is 100 and it fulfills and the condition $f(k)=2^k$ for $k=1, 2, \ldots , 101$ .

Now $f(102)= 2\left({101 \choose 0} +{101 \choose 1}+\ldots +{101 \choose 100}\right) =2\left(2^{101}-1 \right) = 2^{102}-2$ . Since it is given that the last five digits of $2^{101}$ is 10752, then the last five digits of $2^{102}-2$ is 21502.

Note : the question is modified from a question in Pi Mu Epsilon , (1964).