I am thinking about the problem :\begin{equation} \frac{\Gamma(\alpha k+1)}{\Gamma(\alpha k-\alpha+1)} \end{equation}
I know gamma function can represent as a series:\begin{equation} \Gamma(x)=x^{x-1} \sum_{n=1}^{\infty}(-1)^n\binom{x}{n}\sum_{k=1}^n(-1)^k \frac{k !}{k^k}\binom{n}{k} \end{equation} but i stucked as thinking: can we cancel term $k$ to simplify ?
In general, you cannot simplify the expression but you can do is to take logarithms, use twice Stirling approximation, continue with Taylor series, exponentiate.
This would give $$\frac{\Gamma(\alpha k+1)}{\Gamma(\alpha k-\alpha+1)}=(\alpha k)^{\alpha }\sum_{n=0}^\infty\frac{P_n(\alpha)}{c_n\, k^n} $$ where the $c_n$ form sequence $A053657$ in $OEIS$.
The first polynomials $P_n(\alpha)$ are $$\left( \begin{array}{cc} n & P_n(\alpha) \\ 0 & 1 \\ 1 & 1-\alpha \\ 2 & \frac{(\alpha -2) (\alpha -1) (3 \alpha -1)}{\alpha } \\ 3 & -\frac{(\alpha -3) (\alpha -2) (\alpha -1)^2}{\alpha } \\ 4 & \frac{(\alpha -4) (\alpha -3) (\alpha -2) (\alpha -1) \left(15 \alpha ^3-30 \alpha ^2+5 \alpha +2\right)}{\alpha ^3}\\ 5 & -\frac{(\alpha -5) (\alpha -4) (\alpha -3) (\alpha -2) (\alpha -1)^2 \left(3 \alpha ^2-7 \alpha -2\right)}{\alpha ^3}\\ \end{array} \right)$$
Trying for $\alpha=\pi$ and $k=10$, the above truncated formula gives $\color{red}{45245.147961}69$