Proof of $Δv = c \cdot \tanh \left ( \frac{u}{c}\ln \left( \frac{m_0}{m_1}\right) \right) $ 

Question: The relativistic rocket equation is given below 
$$ \frac{m_0}{m_1} = \left (\frac{1+\frac{Δv}{c}}{1-\frac{Δv}{c}} \right)^{\frac{c}{2u}} $$ 
Show that the equation rearranges to 
$$ Δv = c \cdot \tanh \left (  \frac{u}{c}\ln \left( \frac{m_0}{m_1}\right) \right) $$
where the hyperbolic tangent function is $ \tanh(x) = \frac{e^{2x}-1}{e^{2x}+1} $



My attempt:
$$ \frac{m_0}{m_1} = \left (\frac{1+\frac{Δv}{c}}{1-\frac{Δv}{c}} \right)^{\frac{c}{2u}} $$ 
$$ \left( \frac{m_0}{m_1} \right)^{\frac{2u}{c}} = \left (\frac{1+\frac{Δv}{c}}{1-\frac{Δv}{c}} \right)$$ 
$$ \left( \frac{m_0}{m_1} \right)^{\frac{2u}{c}} = \left (\frac{\frac{c+Δv}{c}}{\frac{c-{Δv}}{c}} \right)$$ 
$$ \left( \frac{m_0}{m_1} \right)^{\frac{2u}{c}} = \frac{c+Δv}{c} \cdot \frac{c}{c-Δv} $$
$$ \left( \frac{m_0}{m_1} \right)^{\frac{2u}{c}} = \frac{c+Δv}{c-Δv} $$
$$ \frac{m_0^{\frac{2u}{c}}}{m_1^{\frac{2u}{c}}} = \frac{c+Δv}{c-Δv}  $$
$$ \left(c-Δv\right)m_0^{\frac{2u}{c}} =  \left(c+Δv\right)m_1^{\frac{2u}{c}} $$
$$ cm_0^{\frac{2u}{c}} - Δvm_0^{\frac{2u}{c}} = cm_1^{\frac{2u}{c}} + Δvm_1^{\frac{2u}{c}} $$ 
$$ cm_0^{\frac{2u}{c}} - cm_1^{\frac{2u}{c}} = Δvm_0^{\frac{2u}{c}} + Δvm_1^{\frac{2u}{c}} $$
$$ c(m_0^{\frac{2u}{c}} - m_1^{\frac{2u}{c}}) = Δv(m_0^{\frac{2u}{c}} + m_1^{\frac{2u}{c}})$$
$$ Δv = \frac{ c(m_0^{\frac{2u}{c}} - m_1^{\frac{2u}{c}})}{(m_0^{\frac{2u}{c}} + m_1^{\frac{2u}{c}})} $$
Now I am stuck 
 A: We are given the expression 
$$\frac{m_0}{m_1}=\left(\frac{1+\frac{\Delta v}{c}}{1-\frac{\Delta v}{c}}\right)^{c/2u} \tag 1$$
Raising each side of $(1)$ to the $2u/c$ power yields
$$\left(\frac{m_0}{m_1}\right)^{2u/c}=\frac{1+\frac{\Delta v}{c}}{1-\frac{\Delta v}{c}}$$
whereupon solving for $\Delta v/c$ we find that
$$\frac{\Delta v}{c}=\frac{\left(\frac{m_0}{m_1}\right)^{2u/c}-1}{\left(\frac{m_0}{m_1}\right)^{2u/c}+1} \tag 3$$
Then, multiply the numerator and denominator of $(3)$ by $\left(\frac{m_0}{m_1}\right)^{-u/c}$ to obtain
$$\begin{align}
\frac{\Delta v}{c}&=\frac{\left(\frac{m_0}{m_1}\right)^{u/c}-\left(\frac{m_0}{m_1}\right)^{-u/c}}{\left(\frac{m_0}{m_1}\right)^{u/c}+\left(\frac{m_0}{m_1}\right)^{-u/c}} \tag 4\\\\
&=\frac{e^{(u/c)\log(m_0/m_1)}-e^{(u/c)\log(m_0/m_1)}}{e^{(u/c)\log(m_0/m_1)}+e^{(u/c)\log(m_0/m_1)}} \tag 5\\\\
&=\tanh\left(\frac{u}{c}\log\left(\frac{m_0}{m_1}\right)\right) \tag 6\\\\
\Delta v&=c\,\tanh\left(\frac{u}{c}\log\left(\frac{m_0}{m_1}\right)\right)
\end{align}$$
as was to be shown!

NOTES:
In going from $(4)$ to $(5)$, we made use of the identity $x=e^{\log(x)}$.
In going from $(5)$ to $(6)$, we used the definition of the hyperbolic tangent, $\tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}$

ADDRESSING THE POINT AT WHICH THE OP WAS STUCK
Starting with the expression 
$$Δv = \frac{ c(m_0^{\frac{2u}{c}} - m_1^{\frac{2u}{c}})}{(m_0^{\frac{2u}{c}} + m_1^{\frac{2u}{c}})} \tag 7
$$
we divide numerator and denominator of $(7)$ by $(m_0\,m_1)^{u/c}$ to obtain
$$\frac{\Delta v}{c}=\frac{\left(\frac{m_0}{m_1}\right)^{u/c}-\left(\frac{m_0}{m_1}\right)^{-u/c}}{\left(\frac{m_0}{m_1}\right)^{u/c}+\left(\frac{m_0}{m_1}\right)^{-u/c}} \tag 8$$
Observe that $(8)$ is identical to $(4)$.  Now, finish as in the development that follows $(4)$.
