Let $$b_n=\sum_{i=1}^{n}b_{i-1}b_{n-i}$$ and $b_0$=1.
We are supposed to show that $$B(x)=\sum_{n=0}^{\infty}b_nx^n=xB(x)^2+1$$ I have arrived at the following: $$B(x)=1+x\sum_{n=0}^{\infty}\left(\sum_{i=0}^{n} b_i b_{n-i}\right)x^n$$ My question: how do I get to $$=1+x\left(\sum_{n=0}^{\infty}b_nx^n\right)\left(\sum_{n=0}^{\infty}b_nx^n\right)$$
Using product rule (for series called Cauchy product, you can also verify it by induction)$$(\sum_{i=0}^n a_ix^i)(\sum_{j=0}^n b_jx^j)=\sum_{k=0}^{2n}(\sum_{i=0}^ka_ib_{k-i})x^k$$You can see that $$[B(x)]^2=(\sum_{n=0}^\infty b_n x^n)^2=\sum_{n=0}^\infty (\sum_{k=0}^n b_kb_{n-k})x^n=\sum_{n=0}^\infty b_{n+1}x^n $$ That is what you want