How is it that the equilibrium constant does not depend on the mechanism?
You should indeed be able to write every equilibrium reaction in this simple form. I guess that the reason for this might lie in the principle of microscopic reversibility (shameless self-plug: see this answer of mine). But for non-elementary reactions, i.e. those whose reaction mechanism consists of more than one step, you get a more complicated equilibrium constant that is the product of the equilibrium constants of all the intermediate reaction steps (if the reaction mechanism has no branches). An example: Consider the reaction

\begin{equation}
  \ce{H2(g) + I2(g) &lt;=&gt;&gt; 2HI(g)} \ .
\end{equation}

Its mechanism consists of three reaction steps:


$\ce{ I2 &lt;&lt;=&gt; 2I } \qquad \qquad \qquad \qquad K_{1} = \frac{[\ce{I}]^{2}}{[\ce{I2}]}$
$\ce{I + H2 &lt;&lt;=&gt; H2I} \ \ \quad \qquad \qquad K_{2} = \frac{[\ce{H2I}]}{[\ce{I}] [\ce{H2}]}$
$\ce{H2I + I &lt;=&gt;&gt; 2HI} \quad \qquad \qquad K_{3} = \frac{[\ce{HI}]^{2}}{[\ce{I}] [\ce{H2I}]}$


Now, take a look at the equilibrium constant of the overall reaction

\begin{equation}
  K_{\mathrm{tot}} = \frac{[\ce{HI}]^{2}}{[\ce{I2}] [\ce{H2}]} \ .
\end{equation}

Then make the following substitutions: From the third reaction step's equilibrium equation you get $[\ce{HI}]^{2} = K_{3} [\ce{I}] [\ce{H2I}]$. Substituting this into the equation for $K_{\mathrm{tot}}$

\begin{equation}
  K_{\mathrm{tot}} = \frac{K_{3} [\ce{I}] [\ce{H2I}]}{[\ce{I2}] [\ce{H2}]} \ .
\end{equation}

From reaction step 2 you get $[\ce{H2I}] = K_{2} [\ce{I}] [\ce{H2}]$ which then gives

\begin{equation}
  K_{\mathrm{tot}} = \frac{K_{3} [\ce{I}] K_{2} [\ce{I}] [\ce{H2}] }{[\ce{I2}] [\ce{H2}]} = \frac{K_{3} K_{2} [\ce{I}]^{2}}{[\ce{I2}]} \ .
\end{equation}

And finally, from reaction step 1 you get $[\ce{I}]^{2} = K_{1} [\ce{I2}]$. Substituting this into the equation for $K_{\mathrm{tot}}$ you see that the overall equilibrium constant reduces to the product of the equilibrium constants of the reaction steps

\begin{equation}
  K_{\mathrm{tot}} = \frac{K_{3} [\ce{I}] K_{2} [\ce{I}] [\ce{H2}] }{[\ce{I2}] [\ce{H2}]} = \frac{K_{3} K_{2} K_{1} [\ce{I2}]}{[\ce{I2}]} = K_{1} K_{2} K_{3} \ .
\end{equation}

If you have a reaction that doesn't proceed along a chain of reaction steps but branches out the situation will become more complicated but can be treated in an analogous way as above.