A random variable [tex]X[/tex] following a binomial distribution with success probability [tex]p[/tex] across [tex]n[/tex] trials has PMF
[tex]\mathbb P(X=x)=\begin{cases}\dbinom nxp^x(1-p)^{n-x}&\text{for }x\in\{0,1,\ldots,n\}\\\\0&\text{otherwise}\end{cases}[/tex]
where [tex]\dbinom nx=\dfrac{n!}{x!(n-x)!}[/tex].
The mean of the distribution is given by the expected value which is defined by
[tex]\mathbb E(X):=\displaystyle\sum_xx\mathbb P(X=x)[/tex]
where the summation is carried out over the support of [tex]X[/tex]. So the mean is
[tex]\displaystyle\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}[/tex]
Because this is a proper distribution, you have
[tex]\displaystyle\sum_x\mathbb P(X=x)=1[/tex]
which is a fact that will be used to evaluate the sum above.
[tex]\displaystyle\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}[/tex]
[tex]\displaystyle\sum_{x=1}^nx\binom nxp^x(1-p)^{n-x}[/tex]
[tex]\displaystyle\sum_{x=1}^n\frac{xn!}{x!(n-x)!}p^x(1-p)^{n-x}[/tex]
[tex]\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!(n-x)!}p^{x-1}(1-p)^{n-x}[/tex]
[tex]\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}[/tex]
Letting [tex]y=x-1[/tex], this becomes
[tex]\displaystyle np\sum_{y=0}^{n-1}\frac{(n-1)!}{y!((n-1)-y)!}p^y(1-p)^{(n-1)-y}[/tex]
Observe that the remaining sum corresponds to the PMF of a new random variable [tex]Y[/tex] which also follows a binomial distribution with success probability [tex]p[/tex], but this time across [tex]n-1[/tex] trials. Therefore the sum evaluates to 1, and you're left with [tex]np[/tex] as the expression for the mean for [tex]X[/tex].
