The product of exp is the exponential of a formal power series. This can be expressed using the Bell polynomials:
\( \prod_{n=0}^{\infty} \exp(a_n y^n) = \exp\left(\sum_{n=0}^{\infty} a_n y^n\right) = \exp(a_0) \exp\left(\sum_{n=1}^{\infty} a_n y^n\right) \).
This then becomes
\( \exp(a_0) \exp\left(\sum_{n=1}^{\infty} a_n y^n\right) = \exp(a_0) \sum_{n=1}^{\infty} \frac{\sum_{k=1}^n B_{n,k}(1! a_1, ..., (n-k+1)! a_{n-k+1})}{n!} y^n = \exp(a_0) \sum_{n=1}^{\infty} \frac{B_n(1! a_1, ..., n! a_n)}{n!} y^n \).
Thus the equations to solve are
\( a_n L^n = \exp(a_0) \frac{B_n(1! a_1, ..., n! a_n)}{n!} \).
Since \( a_0 = L \) and \( a_1 = 1 \), this is
\( a_n L^n = L \frac{B_n(1, 2! a_2, ..., n! a_n)}{n!} \).
This is derived from Faà di Bruno's formula, see
http://en.wikipedia.org/wiki/Fa%C3%A0_di...7s_formula
for details.
\( \prod_{n=0}^{\infty} \exp(a_n y^n) = \exp\left(\sum_{n=0}^{\infty} a_n y^n\right) = \exp(a_0) \exp\left(\sum_{n=1}^{\infty} a_n y^n\right) \).
This then becomes
\( \exp(a_0) \exp\left(\sum_{n=1}^{\infty} a_n y^n\right) = \exp(a_0) \sum_{n=1}^{\infty} \frac{\sum_{k=1}^n B_{n,k}(1! a_1, ..., (n-k+1)! a_{n-k+1})}{n!} y^n = \exp(a_0) \sum_{n=1}^{\infty} \frac{B_n(1! a_1, ..., n! a_n)}{n!} y^n \).
Thus the equations to solve are
\( a_n L^n = \exp(a_0) \frac{B_n(1! a_1, ..., n! a_n)}{n!} \).
Since \( a_0 = L \) and \( a_1 = 1 \), this is
\( a_n L^n = L \frac{B_n(1, 2! a_2, ..., n! a_n)}{n!} \).
This is derived from Faà di Bruno's formula, see
http://en.wikipedia.org/wiki/Fa%C3%A0_di...7s_formula
for details.