Using independence, EetΣXi=E"\\Pi" etXi="\\Pi"EetXi=(pet+(1-p))n,
where the Xi are independent Bernoulli random variables. Equivalently
"\\sum^n_{k=0}e^{tk}(^n_k)p^k(1-p)^{n-k}=\\sum^n_{k=0}(^n_k)(pe^{t})^k(1-p)^{n-k}=(pe^t+(1-p)^n)" by the Binomial formula.
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