= Solution
Start from $I_0(\lambda)=1/\lambda$. <Differentiation under the integral sign> gives
$$
I_n'(\lambda)=-I_{n+1}(\lambda),\qquad I_n(\lambda)=(-1)^n\frac{d^n}{d\lambda^n}\frac1\lambda.
$$
This interchange is justified locally around any $\lambda>0$ by the <dominated convergence theorem>: the differentiated integrands are bounded by an integrable <polynomial> times $e^{-\lambda x/2}$. Successive <derivatives> of $\lambda^{-1}$ give
$$
\boxed{I_n(\lambda)=\frac{n!}{\lambda^{n+1}}.}
$$
This is also the integer case of the <Gamma function> integral after scaling the integration variable.
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