How is the equation of Mach number derived?

If you want to know more about calculating a Mach number, it helps to read Wikipedia's article on Mach number. As explained here, the Mach number for subsonic compressible flow is obtainable from Bernoulli's equation (Wikipedia cites this source). The result you cited then follows from $\gamma=\frac{7}{5},\,p_t=q_e+p$.

$M$ is not the speed of sound. It is the Mach number -- the ratio of speed of the aircraft to the speed of sound. The equation is derived from Bernoulli's equation together with a suitable choice of $\gamma=C_P/C_V$ for air.

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.