Let $b_n$ be the number of all unordered binary trees with n leaves. Then it's known: $$b_n = {1\over 2}\sum_{i=1}^{n-1}\binom{n}{i}b_ib_{n-i}$$ Then let $B(x)$ be the $e.g.f $ of $b_n$:
$$B(x) = \sum_{n=0}^\infty b_n\frac{x^n}{n!}\\=\sum_{n=2}^\infty b_n\frac{x^n}{n!}+x\\=\sum_{n=2}^\infty ({1\over 2}\sum_{i=1}^{n-1}\binom{n}{i}b_ib_{n-i})\frac{x^n}{n!}+x$$
From here I would like to induce the well-know formula :
$$B(x)=\frac{1}{2}B(x)^2 -x$$
I think I had found the $-x$ part, however, still missing how to induce square form of $B(x)$