For the binomial distribution, why does no unbiased estimator exist for $1/p$?

Assume that $U:X\mapsto U(X)$ is an unbiased estimator of $1/p$ for some given value $p$ in $(0,1]$. This means that $E_p(U(X))=1/p$, that is, that $G(p)=1$, where $$ G(p)=pE_p(U(X))=\sum_{k=0}^n{n\choose k}U(k)p^{k+1}(1-p)^{n-k}. $$ Since $G$ is a polynomial of degree at most $n+1$, the equation $G(p)=1$ has at most $n+1$ roots. Thus, any estimator $U$ of $1/p$ can be unbiased for at most $n+1$ values of $p$. In particular, no estimator of $1/p$ can be unbiased for every $p$ in $(0,1)$ (the situation the question asks about). Likewise, no estimator of $1/p$ can be unbiased for every $p$ in $(1/2,1)$ (a situation such that $1/p$ is uniformly bounded, as mentioned in the comments).

The same applies to every rational fraction $Q(p)/R(p)$ instead of $1/p$, for some polynomials $Q$ and $R$ such that $Q(0)\ne0=R(0)$.

Note that $$\frac{1}{p} = \frac{1}{1-(1-p)} = 1+(1-p)+\frac{(1-p)^2}{2!}+...$$

Assume there exists unbiased estimator $T(x)$ for $\frac{1}{p}$. Then we have $$\sum_{k=0}^{n} T(k){n \choose k}p^k(1-p)^k = 1+(1-p)+\frac{(1-p)^2}{2!}+... $$

Observe that the LHS is a finite power series, and the RHS is an infinite one. Hence they cannot be equal.