The answer to the first part of your question is $e^2$ and the second part of your question is silly and so I'm not gonna answer that. To show that $\lim_{n\to\infty}b_n=e^2,$ we use this general result that if $\{x_n\}$ is a sequence of real number with $x_n > -1$ and $\lim_{n\to\infty}nx_n=x,$ then $\lim_{n\to\infty}(1+x_n)^n=e^x.$ Now let $$x_n:=\sum_{k=1}^{n-1}\frac{1}{\binom{n}{k}}.$$Then $b_n=a_n^n=(1+x_n)^n$ and $$\lim_{n\to\infty}nx_n=\lim_{n\to\infty}n\sum_{k=1}^{n-1}\frac{1}{\binom{n}{k}}=2+\lim_{n\to\infty}n\sum_{k=3}^{n-3}\frac{1}{\binom{n}{k}}. \ \ \ \ \ \ \ \ \ (*)$$We also have $$0 < n\sum_{k=3}^{n-3}\frac{1}{\binom{n}{k}} < \frac{n(n-5)}{\binom{n}{3}}$$and so, by the squeeze theorem, $\lim_{n\to\infty}n\sum_{k=3}^{n-3}\frac{1}{\binom{n}{k}}=0.$ Thus $\lim_{n\to\infty}nx_n=2,$ by $(*),$ and the result follows.