Conservation of the Laplace–Runge–Lenz vector

Differentiate: $\dfrac{d\mathbf A}{dt} = \ddot{\mathbf r}\times\mathbf h + \dot{\mathbf r}\times\dot{\mathbf h} - GM\dfrac{d\hat{\mathbf r}}{dt}$. Since $\mathbf h$ is constant (Theorem 1), the middle term vanishes. Using $\ddot{\mathbf r}=-\dfrac{GM}{r^{2}}\hat{\mathbf r}$ and $\mathbf h=\mathbf r\times\dot{\mathbf r}$, the vector triple product identity $\hat{\mathbf r}\times(\mathbf r\times\dot{\mathbf r}) = \mathbf r(\hat{\mathbf r}\cdot\dot{\mathbf r}) - \dot{\mathbf r}(\hat{\mathbf r}\cdot\mathbf r)$ gives $\ddot{\mathbf r}\times\mathbf h = -\dfrac{GM}{r^{2}}\big[r\dot r\,\hat{\mathbf r} - r\dot{\mathbf r}\big] = GM\left(\dfrac{\dot{\mathbf r}}{r} - \dfrac{\dot r}{r}\hat{\mathbf r}\right)$, where $\dot r=\hat{\mathbf r}\cdot\dot{\mathbf r}$ is the scalar rate of change of the distance $r$ (used $\hat{\mathbf r}\cdot\mathbf r=r$).

On the other hand, differentiating $\hat{\mathbf r}=\mathbf r/r$ directly gives $\dfrac{d\hat{\mathbf r}}{dt} = \dfrac{\dot{\mathbf r}}{r} - \dfrac{\mathbf r\,\dot r}{r^{2}} = \dfrac{\dot{\mathbf r}}{r} - \dfrac{\dot r}{r}\hat{\mathbf r}$ — exactly the same bracketed expression found for $\ddot{\mathbf r}\times\mathbf h/GM$ above. So $\ddot{\mathbf r}\times\mathbf h = GM\dfrac{d\hat{\mathbf r}}{dt}$ identically.

Therefore $\dfrac{d\mathbf A}{dt} = GM\dfrac{d\hat{\mathbf r}}{dt} - GM\dfrac{d\hat{\mathbf r}}{dt} = \mathbf 0$, proving $\mathbf A$ is constant. Evaluating $\mathbf A$ at periapsis, where $\dot{\mathbf r}$ is purely tangential (perpendicular to $\hat{\mathbf r}$) with speed $v_p=h/r_p$, gives $\dot{\mathbf r}\times\mathbf h$ pointing along $-\hat{\mathbf r}$ with magnitude $h^{2}/r_p$, so $\mathbf A = \left(\dfrac{h^{2}}{r_p}-GM\right)\hat{\mathbf r}$; using $r_p=p/(1+e)=\dfrac{h^{2}}{GM(1+e)}$ from the orbit equation gives $\mathbf A = GMe\,\hat{\mathbf r}_p$, confirming $|\mathbf A|=GMe$ pointing exactly toward periapsis. $\blacksquare$