= Solution
Adding and subtracting the canonical <scalar field> density and pressure gives
$$
\dot\phi^2=(1+w)\rho_\phi,\qquad V=\frac{1-w}{2}\rho_\phi.
$$
A real scalar with positive kinetic energy consequently needs $w\geq-1$. On the expanding scalar-dominated branch, divide $\dot\phi$ by $H$ and use the <Friedmann equation>:
$$
\frac{d\phi}{d\log a}=\frac{\dot\phi}{H}
=\pm\sqrt{\frac{3(1+w)}{8\pi G}}.
$$
Choose the positive rolling sign. Integration and elimination of $a$ using $\rho_\phi=\rho^0_\phi(a/a_0)^{-3(1+w)}$ give the <scalar-field reconstruction for a constant equation of state>:
$$
\boxed{\phi(a)=\phi_0+\sqrt{\frac{3(1+w)}{8\pi G}}\log\frac a{a_0},}
$$
$$
\boxed{V(\phi)=\frac{1-w}{2}\rho^0_\phi
\exp\left[-\sqrt{24\pi G(1+w)}(\phi-\phi_0)\right].}
$$
Changing the rolling sign reverses the exponential slope. At $w=-1$, $\phi$ is constant and $V=\rho^0_\phi$; elimination by an invertible $\phi(a)$ then degenerates. The <cosmological continuity equation> gives $\ddot\phi+3H\dot\phi+V_{,\phi}=0$ whenever $\dot\phi\ne0$, so the reconstruction satisfies the field equation as well as the Friedmann equation.
Compare the reconstructed exponent with $V_0e^{-\lambda\sqrt{8\pi G}\phi}$. In this constant-$w$, scalar-dominated scaling solution,
$$
\lambda^2=3(1+w),\qquad w=-1+\frac{\lambda^2}{3},\qquad a\propto t^{2/\lambda^2}.
$$
Strict $\rho_\phi+P_\phi>0$ requires $w>-1$, and acceleration requires $w<-1/3$. Hence
$$
\boxed{0<\lambda<\sqrt2.}
$$
The accelerating branch has $V_0>0$. The zero-slope limit is a constant potential with $\rho+P=0$, so it does not meet the strict inequality. This result characterizes the <exponential-potential power-law inflation> solution; an arbitrary trajectory with the same potential need not have constant $w$ throughout its evolution.
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