Let's limit the flexible polynomial

We can express $P(x)=1729\displaystyle\prod_{i=1}^{n}(x-\alpha_i)$ and we can write $L$ as $\displaystyle\lim_{x->\alpha_1}\left(1+1729\displaystyle\prod_{i=1}^{n}(x-\alpha_i)\right)^{\frac{1729\displaystyle\prod_{i=2}^{n}(x-\alpha_i)}{1729\displaystyle\prod_{i=1}^{n}(x-\alpha_i)}}$ . Note that as $x\rightarrow \alpha_1$ , $1729\displaystyle\prod_{i=1}^{n}(x-\alpha_i) \rightarrow 0$ , so the limit $L$ is equivalent to the famous limit $\displaystyle\lim_{x\rightarrow 0} (1+x)^{1/x} = e$ . So finally, we have:

$L=e^{\displaystyle\lim_{x\rightarrow \alpha_1} 1729\displaystyle\prod_{i=2}^{n}(x-\alpha_i)}=e^{1729k}$ and we have $\dfrac{\ln L}{k}=\boxed{1729}$ .