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