The proof is based on the Binomial Theorem for the expansion of
(1+x)r.
where for each value of c,
the inner sum in the last sum is taken over all sets
such that
and
.
But there is at most one such
set of coefficients, given by sm = cm
if every cm <= rm (since there
is a unique representation for the p-ary form of c.) If cm >
rm for some m, then the inner sum is zero. In either case, the
theorem follows by equating coefficients of xc for each 0
<= c <= r.
Back to the mathematical behavior of
Pascal's Triangle (mod 2).