It seems the best way to describe the moments of $X_i$ is to use ascending factorials. Using Pochhammer's notation $ (x)_p=x(x+1)\cdots(x+p-1) $ for every nonnegative integer $p$, one gets $ E((X_i)_p)=(i)_p\frac{(m+1)_p}{(n+1)_p}. $ Since $(x)_1=x$, this yields your formula for $E(X_i)$. Using the shorthand $ a_p=(i+p-1)\frac{m+p}{n+p}, $ one gets $E((X_i)_p)=a_1a_2\cdots a_{p}$. This yields (I think) the value of the kurtosis of $X_i$ defined by $ \beta_2(X_i)=\frac{E((X_i-E(X_i))^4)}{E((X_i-E(X_i))^2)^2} $ as the ratio $ \beta_2(X_i)=\frac{a_2a_3a_4-6a_2a_3+7a_2-1-4a_1(a_2a_3-3a_2+1)+6a_1^2(a_2-1)-3a_1^3}{a_1(a_2-a_1-1)^2}. $ And I am well aware that this solution does not use Maple...
Edit To get the formula for $E((X_i)_p)$ written above, one can denote by $X_i^{m,n}$ the random variable considered by the OP and sum over every possible value of $k$ the algebraic identity $ P(X_i^{m,n}=k)\cdot (k)_p=P(X_{i+p}^{m+p,n+p}=k+p)\cdot (i)_p\frac{(m+1)_p}{(n+1)_p}. $ The LHS of the resulting equality is $E((X_i^{m,n})_p)$ and the RHS is the total mass of the distribution of $X_{i+p}^{m+p,n+p}$, which equals $1$, times the factor $(i)_p(m+1)_p/(n+1)_p$. This proves the formula.