= Solution
Assume a common support, a differentiable likelihood, and dominated differentiation under the integrals. The <score function> $S_\theta=\partial_\theta\log f(X;\theta)$ has mean zero. Its second moment is the <Fisher information> $I(\theta)$. If $T$ is unbiased, differentiation of $\mathbb E_\theta T=\theta$ gives $\mathbb E_\theta(TS_\theta)=1$, hence $\operatorname{Cov}_\theta(T,S_\theta)=1$. The <Cauchy-Schwarz inequality> yields
$$
\boxed{\operatorname{Var}_\theta(T)\geq I(\theta)^{-1},}
$$
the <Cramér-Rao lower bound>, assuming positive finite information.
Equality holds precisely when the centred estimator and score are linearly dependent almost surely. The covariance fixes the factor, giving $S_\theta=I(\theta)(T-\theta)$. If the same unbiased estimator is efficient for every parameter on a regular interval, integrate this identity in $\theta$:
$$
\log f(x;\theta)=A(\theta)T(x)-B(\theta)+C(x),
\qquad A'=I,\quad B'=\theta I.
$$
Exponentiation gives $f(x;\theta)=h(x)\exp(A(\theta)T(x)-B(\theta))$, an <exponential family>. This proves <efficient unbiased estimator implies exponential family> under the stated common-support regularity, rather than asserting it for nonregular support-dependent models.
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