I am trying to show that the likelihood equation of the following pmf with $\theta\in(0,\infty)$ and $X\in(1,2,3...)$ has one unique root:
$$P(X_i=x)=\frac{\theta^x e^{-\theta}}{x!(1-e^{-\theta})}$$
The likelihood equation(a function of $\theta$) is (for a fixed $x$) :
$$\frac{\partial}{\partial\theta}\log(P(X_i=x)=\frac{\partial}{\partial\theta}log\bigg[\frac{\theta^x e^{-\theta}}{x!(1-e^{-\theta})}\bigg]=\frac{x}{\theta}-\frac{1}{1-e^{-\theta}}$$
Upon combining them:
$$\frac{\partial}{\partial\theta}\log(P(X_i=x)=\frac{x(1-e^{-\theta})-\theta}{\theta(1-e^{-\theta})}$$
Since the denominator can never be zero, the roots of the likelihood equation are equivalent to the roots of $x(1-e^{-\theta})-\theta$.
Consider:
$$\frac{\partial}{\partial\theta}\bigg[x(1-e^{-\theta})-\theta\bigg]=xe^{-\theta}-1$$
With the unique solution of $\theta=\ln(x)$. This implies if roots exist then can only have two roots for any fixed $x$.
Note, if $\theta=0$ then $x(1-e^{-\theta})-\theta=0$, so our function is concave and because we are only considering $\theta\in(0,\infty)$ then $x(1-e^{-\theta})-\theta$ has one unique root and by equivalency so does the likelihood equation.
Is my logic correct? Is there a simpler way to solve this problem?