Using the recurrence relation $2\beta_{n+1} = \sum_{k=0}^{n} \binom{n}{k} \beta_k \beta_{n-k}$ and the analogous relation $2\alpha_{n+1} = \sum_{k=0}^{n} \binom{n}{k} \alpha_k \alpha_{n-k}$ with $\alpha_0 = \beta_0 = 1$, deduce that $\beta_n = \alpha_n$ for every $n \in \mathbb{N}$.