 Zero-truncated binomial distribution

The zero-truncated binomial distribution has probabilities $\left(\genfrac{}{}{0ex}{}{n}{x}\right)p^x(1-p{)}^{n-x}/[1-(1-p{)}^n]$ for $1\le x\le n$. Its natural parameter is $\varphi =\log [p/(1-p)]$ and its cumulant is $\log [(1+e^{\varphi }{)}^n-1]$. Mean and variance are $np/s$ and $np(1-p)/s-n^2{p}^2(1-p{)}^n/s^2$, where $s=1-(1-p{)}^n$. The conditional normalizer depends on $p$, so its likelihood is not proportional to the untruncated likelihood even after a positive observation. At $X=1$ and $n>1$, its likelihood supremum is at $p\downarrow 0$ rather than at $1/n$.