Show that for all $k \in \mathbf { N } ^ { * }$, $$g ^ { ( k ) } ( 0 ) = ( - 1 ) ^ { k } d _ { k }$$ where the coefficients $d _ { k }$ are defined by $$d _ { 0 } = 1 , \quad \text { and } \quad \forall k \geq 1 \quad d _ { k } = \sum _ { i = 1 } ^ { k } \binom { k - 1 } { i - 1 } d _ { k - i } b _ { i } ,$$ and $g \in \mathcal { C } ^ { \infty } \left( \mathbf { R } _ { + } , \mathbf { R } \right)$ is defined by $g ( x ) = \mathrm { e } ^ { y ( x ) }$, with $y(x) = \sum_{n=0}^{+\infty} a_n e^{-\lambda_n x}$ and $b_k = \sum_{n=1}^{+\infty} \lambda_n^k a_n$.
Show that for all $k \in \mathbf { N } ^ { * }$,
$$g ^ { ( k ) } ( 0 ) = ( - 1 ) ^ { k } d _ { k }$$
where the coefficients $d _ { k }$ are defined by
$$d _ { 0 } = 1 , \quad \text { and } \quad \forall k \geq 1 \quad d _ { k } = \sum _ { i = 1 } ^ { k } \binom { k - 1 } { i - 1 } d _ { k - i } b _ { i } ,$$
and $g \in \mathcal { C } ^ { \infty } \left( \mathbf { R } _ { + } , \mathbf { R } \right)$ is defined by $g ( x ) = \mathrm { e } ^ { y ( x ) }$, with $y(x) = \sum_{n=0}^{+\infty} a_n e^{-\lambda_n x}$ and $b_k = \sum_{n=1}^{+\infty} \lambda_n^k a_n$.