This answer is mistaken, as noted in comments ...
That's a nice, simple representation; I wasn't aware of it, but presumably something this simple must be known. Here's a proof:
The case $i+j = n$ is sufficient to get degree $n$ polynomials (perhaps that's what you meant).
Here is a plot of this positive basis, for $n = 3$; the picture suggests a working strategy.
alt text http://dl.dropbox.com/u/5390048/PositivePolynomials.jpg
Since only one of the functions is positive at each endpoint, you know what the coefficients of these functions must be. If you subtract, is it still positive? That follows from
Lemma: any degree $n$ polynomial positive in the unit interval that takes value 1 at 1 must be greater than $x^n$.
(My proof was fallacious ).
From the lemma, it follows that $q(x) = p(x) - p(0) (1-x)^n + p(1) x^n$ is still positive. Since $q(x)$ is 0 at the endpoints of the interval, it is divisible by $x(1-x)$. Use induction to represent the quotient; this gives the desired representation for $p$.