Trivially, by construction you get by summing over \(R = \{1, 2, ... , n \}\)
\begin{equation*}
\sum_{x=1}^n \frac{1}{n} = 1
\end{equation*}
Also, 1/n is positive for all x values.
To determine the mean 1,
\begin{align*}
\mu & = \sum_{x=1}^n x \cdot \frac{1}{n}\\
& = \frac{1}{n}\sum_{x=1}^n x \\
& = \frac{1}{n} \frac{n(n+1)}{2}\\
& = \frac{1+n}{2}
\end{align*}
To determine the variance 2,
\begin{align*}
\sigma^2 & = \sum_{x=1}^n x^2 \cdot \frac{1}{n} - \mu^2\\
& = \frac{1}{n}\sum_{x=1}^n x^2 - \left ( \frac{1+n}{2}\right )^2 \\
& = \frac{1}{n} \frac{n(n+1)(2n+1)}{6} - \frac{1+2n+n^2}{4}\\
& = \frac{(2n^2+3n+1)}{6} - \frac{1+2n+n^2}{4}\\
& = \frac{(4n^2+6n+2)}{12} - \frac{3+6n+3n^2}{12}\\
& = \frac{(n^2-1)}{12}
\end{align*}
For skewness 3,
\begin{align*}
\gamma_1 = & \sum_{x=1}^n x^3 \cdot \frac{1}{n} - 3 \mu \sum_{x=1}^n x^2 \cdot \frac{1}{n} + 2\mu^3\\
& = \frac{n^2(n+1)^2}{4n} - 3\frac{(n(n+1)(2n+1))}{2n} \frac{1+n}{2} + 2 \left ( \frac{1+n}{2}\right )^3 \\
& = \frac{n^2(n+1)^2}{4n} - \frac{(n+1)^2 (n(2n+1)}{4n} + \frac{(n+1)^3}{4}\\
& = \frac{(n+1)^2}{4} \left [ n - 2n -1 + (n+1) \right ]\\
& = 0
\end{align*}
which should be obvious since the histogram for this distribution is constantly flat.
For kurtosis 4, use the fourth moment and simplify. This this is tedious, the algebra is performed using Sage in the active cell below this proof. However, you might want to supply the remainder of this proof using the fact that
\begin{equation*}
\sum_{x=1}^n x^4 = \frac{6n^5 + 15n^4 + 10n^3 - n}{30}.
\end{equation*}