$a_1=1, a_2=1, a_3=3, a_4=15, a_5=105$

Reccurence formula is

$a_{k+1}=\sum\limits_{\lambda_1+\lambda_2+\ldots+\lambda_s=k,\ \lambda_i\geq1} a_{\lambda_1}a_{\lambda_2}...a_{\lambda_s}{k \choose \lambda_1\lambda_2...\lambda_s}$ (1)

($\lambda_1,\lambda_2,\ldots,\lambda_s$ are not ordered)

I need to prove that $a_k=(2k-3)!!$

I've represented (1) into this form

$a_{k+1}\frac{t^k}{k!}=\sum\limits_{\lambda_1+\lambda_2+\ldots+\lambda_s=k,\ \lambda_i\geq1} \prod \limits_{i=1} a_{\lambda_i} \frac{t^{\lambda_i}}{\lambda_i!}$ (2)

I think that after summing by k left side of (2) gives $A'(t)$, where $A(t)=\sum a_i \frac{t^i}{i!}$ (exponential generating function)

but what will be the rightt side of (2)?

is it good to use exponential g.f. here?