special relativity - Relativistic factor between coordinate acceleration and proper acceleration

I did a recent question about relativistic kinematics here: Generalizing a relativistic kinematics formula for spatial-acceleration dependence.

I have a confusion. In the textbooks I've seen, they put the relationship between proper acceleration and coordinate acceleration as

$$ \alpha = \gamma^3 \frac{dv}{dt} $$

However, when I try to do the derivation myself, I get a factor of $\gamma^4$ instead. I'm not sure where the error is.

My derivation is like this:

$$\frac{dx}{d \tau} = \frac{dx}{dt} \frac{dt}{d \tau} = \gamma \frac{dx}{dt}$$

where we used $\frac{dt}{d \tau} = \gamma $

Now,

$$\frac{d^2 x}{d \tau^2} = \frac{ d}{d \tau}( \gamma \frac{dx}{dt}) = \frac{d \gamma}{d \tau} \frac{dx}{dt} + \gamma^2 \frac{ d^2 x}{dt^2} $$

$$\frac{d \gamma}{d \tau} = \gamma \frac{d \gamma}{dt} = \gamma (\frac{ \gamma^3 }{c^2} \frac{dx}{dt} \frac{d^2 x}{dt^2})$$

$$\frac{d^2 x}{d \tau^2} = \gamma^4 \frac{v^2}{c^2} \frac{d^2 x}{dt^2} + \gamma^2 \frac{d^2 x}{dt^2} = \gamma^2 \frac{d^2 x}{dt^2} (\gamma^2 \frac{v^2}{c^2} + 1) = \gamma^4 \frac{d^2 x}{dt^2}$$

where on the last step, $v= \frac{dx}{dt}$ .

I don't understand where is the mistake.