To prove ‾P (1) = p_{n}
Solution: since in the above case we determine each term excluding B_{n,n} (u) will have numerous of (1 - u)^{i} (i = 0 to n) consequently by using u = 1 will lead to outcome = 0 of all terms except of B_{n, n} (u).
B_{n,n}(u) = n!(n!(n-n)!). u^{n }.(1 -u)^{n - n} = u^{n}
P(u - 1) = p_{0}.0 + p1.0 + .............+ p_{n}.1^{n}
= p_{n}