Derivation

Let $\xi_k,$ $k=1,2,\ldots,n,$ be the result of $k$-th experiment, i.e. $\xi_k=1$ if $k$-th experiment was successfull, and $\xi_k=0$ otherwise. Then $$X=\xi_1+\xi_2+\ldots+\xi_n,$$ By assumption, $\xi_1,\ldots,\xi_n$ are independent and each has a Bernoulli distribution with parameter $p.$ Event $\{X=k\}$ means that exactly $k$ variables of $\xi_1\ldots,\xi_n$ equal to $1$ and others are equal to $0.$ There are ${n\choose k}$ possibilities to choose variables that are equal to $1.$ Each of them is $1$ with probability $p,$ other $n-k$ variables are $0$ each with probability $1-p.$

Moment generating function: $$M(t)=(1-p+pe^t)^n$$

Proof

$$M(t)=Ee^{t(\xi_1+\ldots+\xi_n)}=$$ using independence $$=Ee^{t\xi_1}Ee^{t\xi_2}\ldots Ee^{t\xi_n}=(pe^t+1-p)^n$$

Expectation: $EX=np$

Variance: $V(X)=np(1-p)$

Derivation

Expectation is the first derivative $M'(0).$ We have $$M'(t)=n(pe^t+1-p)^{n-1}pe^t, \ EX=M'(0)=np.$$ Second moment is the second derivative $M''(0).$ We have $$M''(t)=n(n-1)(pe^t+1-p)^{n-1}p^2e^{2t}+n(pe^t+1-p)^{n-1}pe^t,$$ $$EX^2=M''(0)=n(n-1)p^2+np.$$ Variance is $$V(X)=EX^2-(EX)^2=n(n-1)p^2+np-n^2p^2=np(1-p)$$