Solution
= Solution
Here $q_{n+1}\ge a_{n+1}q_n=q_n^{n+1}$, so $|\alpha-p_n/q_n|<q_n^{-(n+2)}$. If $\alpha$ were algebraic of fixed degree $d$, Liouville's theorem would contradict this for $n>d$. Thus $\alpha$ is transcendental.
= Solution
Here $q_{n+1}\ge a_{n+1}q_n=q_n^{n+1}$, so $|\alpha-p_n/q_n|<q_n^{-(n+2)}$. If $\alpha$ were algebraic of fixed degree $d$, Liouville's theorem would contradict this for $n>d$. Thus $\alpha$ is transcendental.