I am looking to show that for any real number $\lambda>1$ and integer $n\ge 1$, there are $n$ solutions in the unit disk $z\in\mathbb{C}, |z|\le 1$ to the equation $$z^n e^{-z}=e^{-\lambda}$$ with exactly one being real and positive. I am able to prove that there is indeed exactly one solution that is real and positive. In this case, $z=r, r>0$ and we are looking to solve $$r^ne^{-r}=e^{-\lambda}.$$ We notice that the function $f(r)=r^n e^{-r}$ is continuous on the interval $[0,1]$, taking the values $f(0)=0<e^{-\lambda}$, $f(1)=e^{-r}>e^{-\lambda}$ and $f'(r)>0$ for $r\in[0,1)$. Hence, by the intermediate value theorem, there is one and only one solution to $f(r)=e^{-\lambda}$ for $0\le r\le 1$.
However, I am not quite sure how to show the existence of the $n-1$ other solutions in the unit disk.
Let $\varepsilon(z)=-e^z$ and $f(z)=z^ne^\lambda$. If $|z|=1$, then$$|\varepsilon(z)|\leqslant e<e^\lambda=|f(z)|.$$Therefore, by Rouché's theorem, $f$ and $f+\varepsilon$ have the same number of zeros (counted with their multiplicities) in the open unit disk. But $f$ has $n$ zeros there, and therefore the equation$$f(z)+\varepsilon(z)=0(\iff z^ne^\lambda=e^z\iff z^ne^{-z}=e^{-\lambda})$$has $n$ solutions there.