Expectation of a Binomial Random Variable

Write $X = X_1 + X_2 + \cdots + X_n$, where $X_i$ is $1$ if the $i$-th trial succeeds and $0$ otherwise. Each $X_i$ is a Bernoulli random variable with $P(X_i=1)=p$ and $P(X_i=0)=1-p$, so its expectation is $E[X_i] = 1\cdot p + 0\cdot(1-p) = p$.

Expectation is linear for any random variables, whether or not they are independent: $E[X_1+X_2+\cdots+X_n] = E[X_1]+E[X_2]+\cdots+E[X_n]$. This linearity follows directly from the definition of expectation as a weighted sum, since sums and weighted sums can always be reordered.

Applying linearity to $X = X_1+\cdots+X_n$ gives $E[X] = E[X_1]+\cdots+E[X_n] = \underbrace{p+p+\cdots+p}_{n \text{ times}} = np$, which is exactly the claimed formula.