For the mean 1,
\begin{align*}
\mu & = E[X] \\
& = \sum_{x=0}^{n} {x \binom{n}{x} p^x (1-p)^{n-x}}\\
& = \sum_{x=1}^{n} {x \frac{n(n-1)!}{x(x-1)!(n-x)!} p^x (1-p)^{n-x}}\\
& = np \sum_{x=1}^{n} {\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!} p^{x-1} (1-p)^{(n-1)-(x-1)}}
\end{align*}
Using the change of variables \(k=x-1\) and \(m = n-1\) yields a binomial series
\begin{align*}
& = np \sum_{k=0}^{m} {\frac{m!}{k!(m-k)!} p^k (1-p)^{m-k}}\\
& = np (p + (1-p))^m = np
\end{align*}
For the variance 2,
\begin{gather*}
\sigma^2 = E[X(X-1)] + \mu - \mu^2 \\
= \sum_{x=0}^{n} {x(x-1) \binom{n}{x} p^x (1-p)^{n-x}} + np - n^2p^2\\
= \sum_{x=2}^{n} {x(x-1) \frac{n(n-1)(n-2)!}{x(x-1)(x-2)!(n-x)!} p^x (1-p)^{n-x}} + np - n^2p^2\\
= n(n-1)p^2 \sum_{x=2}^{n} {\frac{(n-2)!}{(x-2)!((n-2)-(x-2))!} p^{x-2} (1-p)^{(n-2)-(x-2)}} + np - n^2p^2
\end{gather*}
Using the change of variables \(k=x-2\) and \(m = n-2\) yields a binomial series
\begin{gather*}
= n(n-1)p^2 \sum_{k=0}^{m} {\frac{m!}{k!(m-k)!} p^k (1-p)^{m-k}} + np - n^2p^2\\
= n(n-1)p^2 + np - n^2p^2 = np - np^2 = np(1-p)
\end{gather*}
The skewness 3 and kurtosis 4 can be found similarly using formulas involving E[X(X-1)(X-2)] and E[X(X-1)(X-2)(X-3)]. The complete determination is performed using Sage below.