The factors are not obvious, I agree.
For instance, for a polytrope index, $\gamma$, of 7/5 the exponent of 2/7 corresponds to a term of the form $\left( \tfrac{\gamma - 1}{\gamma} \right)$, which is our first hint. The second hint is that the pitot tube system can be applied to a Bernoulli system. The third thing to note is that for subsonic speeds, which is where a pitot tube actually functions, one can get away with assuming incompressible flow (I know it seems odd since things obviously do compress a little, but the effects can be considered secondary for most intents and purposes).
For a polytropic ideal gas, we know that $P \propto \rho^{\gamma}$. Thus, we can say that:
$$
P = \kappa \ P_{s} \ \rho^{\gamma} \tag{1}
$$
where $P_{s}$ is the static pressure (also can be considered the pressure at infinity). We can rewrite this equation in terms of density to find:
$$
\rho = \kappa^{-\frac{1}{\gamma}} \ \left( \frac{P}{P_{s}} \right)^{\frac{1}{\gamma}} \tag{2}
$$
The differential form of Bernoulli's equation can be given as:
$$
u \ du + \frac{ 1 }{ \rho } \ \frac{ d P }{ d \rho } \ d \rho = 0 \tag{3}
$$
and we know that the speed of sound is given by:
$$
\begin{align}
C_{s}^{2} & = \frac{ \partial P }{ \partial \rho } \tag{4a} \\
& = \gamma \ \kappa \ P_{s} \ \rho^{\gamma - 1} \tag{4b} \\
& = \frac{ \gamma \ P }{ \rho } \tag{4c}
\end{align}
$$
If we replace the $\rho$ in Equation 4b with the form shown in Equation 2, one can show that the 2nd term in Equation 3 can be rewritten as:
$$
\begin{align}
\frac{ 1 }{ \rho } \ \frac{ d P }{ d \rho } \ d \rho & = \frac{ \gamma \ \kappa \ P_{s} }{ \rho } \ \rho^{\gamma - 1} \ d \rho \tag{5a} \\
& = \frac{ \gamma \ \kappa \ P_{s} }{ \rho } \ \kappa^{-\frac{ \gamma - 1 }{ \gamma }} \ \left( \frac{ P }{ P_{s} } \right)^{\frac{ \gamma - 1 }{ \gamma }} \ d \rho \tag{5b} \\
& = \frac{ \gamma \ \ P_{s} }{ \rho } \ \kappa^{\frac{ 1 }{ \gamma }} \ \left( \frac{ P }{ P_{s} } \right)^{\frac{ \gamma - 1 }{ \gamma }} \ d \rho \tag{5c}
\end{align}
$$
If we differentiate Equation 2, we find:
$$
d \rho = \left( \frac{ \rho }{ \gamma \ P_{s} } \right) \ \left( \frac{ P }{ P_{s} } \right)^{-1} \ dP \tag{6}
$$
so that Equation 5c can be rewritten as:
$$
\frac{ \gamma \ \ P_{s} }{ \rho } \ \kappa^{\frac{ 1 }{ \gamma }} \ \left( \frac{ P }{ P_{s} } \right)^{\frac{ \gamma - 1 }{ \gamma }} \ d \rho = \kappa^{\frac{ 1 }{ \gamma }} \ \left( \frac{ P }{ P_{s} } \right)^{-\frac{ 1 }{ \gamma }} \ dP \tag{7}
$$
We define $u \ du \rightarrow C_{s}^{2} \ M \ dM$, thus we rewrite Equation 3 as:
$$
C_{s}^{2} \ M \ dM + \kappa^{\frac{ 1 }{ \gamma }} \ \left( \frac{ P }{ P_{s} } \right)^{-\frac{ 1 }{ \gamma }} \ dP = 0 \tag{8}
$$
We also define $\alpha = \tfrac{ P }{ P_{s} }$ so that $dP \rightarrow P_{s} \ d\alpha$. If we integrate Equation 8 with the limits ranging from $P_{s}$ to $P$, the change of variables makes the 2nd term go to:
$$
\kappa^{\frac{ 1 }{ \gamma }} \ P_{s} \int_{\alpha}^{1} \ d\alpha \ \alpha^{-\frac{ 1 }{ \gamma }} = \left[ \frac{ \gamma \ \kappa^{\frac{ 1 }{ \gamma }} \ P_{s} }{ \gamma - 1 } \ \alpha^{\frac{ \gamma - 1 }{ \gamma }} \right]_{\alpha}^{1} \tag{9}
$$
Thus, Equation 8 can be rewritten as:
$$
0 = \frac{ 1 }{ 2 } C_{s}^{2} \ M^{2} - \frac{ \gamma \ \kappa^{\frac{ 1 }{ \gamma }} \ P_{s} }{ \gamma - 1 } \left[ \alpha^{\frac{ \gamma - 1 }{ \gamma }} - 1\right] \tag{10}
$$
which after some algebra reduces to:
$$
M^{2} = \frac{ 2 }{ \gamma - 1 } \left[ \alpha^{\frac{ \gamma - 1 }{ \gamma }} - 1\right] \tag{11}
$$
As stated above, for $\gamma$ = 7/5, this results in the form about which you are concerned.
