I Legendre Transform: Momentum & Velocity

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I apologize for the simplicity of the question. I have been reading a paper on the Legendre transform (https://arxiv.org/pdf/0806.1147.pdf), and I am not understanding a particular step in the discussion.

In the paper, Equation 16, where ##\mathcal{H} = \sqrt{\vec{p}^2 + m^2} ##:
\begin{equation}
\begin{split}
\vec{v} &\doteq \frac{d\mathcal{H}}{d \vec{p}}
\\
&= \frac{\vec{p}}{\sqrt{\vec{p}^2 + m^2}}
\end{split}
\end{equation}

Such that, Equation 17
\begin{equation}
\begin{split}
\vec{p} &= m \vec{v} \left(\frac{1}{\sqrt{1-\vec{v}^2}} \right)
\end{split}
\end{equation}

However, I don't understand the derivation of Equation 17 from Equation 16. I assume I am missing something very simple, but I am not seeing it.
 
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Square (16) and solve for ##\vec p^2## in terms of ##\vec v##. Reinsert in the denominator and solve for ##\vec p##.
 
Nice; simple; I figured; thanks!
 
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