Limit with summations

Relevant wiki: Riemann Sums

$\begin{aligned} L & = \lim_{n \to \infty} \lim_{m \to \infty} \sum_{r=1}^n \sum_{k = 1}^{mr} \frac {mn^2}{(m^2n^2+k^2)(n^2+r^2)} & \small \color{#3D99F6} \text{Dividing up and down by }m^2n^2 \\ & = \lim_{n \to \infty} \lim_{m \to \infty} \sum_{r=1}^n \frac 1{n^2+r^2} \cdot \frac 1m \sum_{k = 1}^{mr} \frac 1{1+\left(\frac k{mn}\right)^2} & \small \color{#3D99F6} \text{Using Riemann sums} \\ & = \lim_{n \to \infty} \sum_{r=1}^n \frac 1{n^2+r^2} \int_0^r \frac 1{1+\frac {x^2}{n^2}}\ dx & \small \color{#3D99F6} \lim_{n \to \infty} \frac 1n \sum_{k = a}^b f\left(\frac kn\right) = \lim_{n \to \infty} \int_\frac an^\frac bn f(x) \ dx \\ & = \lim_{n \to \infty} \sum_{r=1}^n \frac 1{n^2+r^2} \cdot n\tan^{-1} \frac xn \ \bigg|_0^r \\ & = \lim_{n \to \infty} \sum_{r=1}^n \frac {n \tan^{-1} \frac rn}{n^2+r^2} & \small \color{#3D99F6} \text{Dividing up and down by }n^2 \\ & = \lim_{n \to \infty} \frac 1n \sum_{r=1}^n \frac {\tan^{-1} \frac rn}{1+\frac {r^2}{n^2}} & \small \color{#3D99F6} \text{Using Riemann sums again} \\ & = \int_0^1 \frac {\tan^{-1} x}{1+x^2} \ dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = \left(\tan^{-1} x\right)^2 \ \bigg|_0^1 - \color{#3D99F6} \int_0^1 \frac {\tan^{-1} x}{1+x^2} \ dx & \small \color{#3D99F6} \text{Note that } \int_0^1 \frac {\tan^{-1} x}{1+x^2} \ dx = L \\ & = \frac 12 \left(\tan^{-1} x\right)^2 \ \bigg|_0^1 \\ & = \frac {\pi^2}{32} \end{aligned}$

Therefore, $\gamma^{\beta^\alpha} = 32^{2^1} = \boxed{1024}$ .