give me a path to do the last part
I get for the log likelihood function
$\displaystyle \ln L=-n\ln(1-e^{-\lambda})-n\lambda+\sum X_i\ln\lambda-\ln(\prod (X_i!))$
The derivative wrt lambda is
$\displaystyle {-ne^{-\lambda} \over 1-e^{-\lambda}}-n+{\sum X_i\over \lambda}$
set this equal to zero and divide by n ...
$\displaystyle {\bar X\over \lambda}={1 \over 1-e^{-\lambda}}$
I don't think we can find the MLE of $\displaystyle \lambda$ say $\displaystyle \hat \lambda$
But I don't think they want you to either.