Electric field inside a conductor

I would like to clear things up: How exactly the electric field inside a conductor is zero? Let a really "powerful" electric field be outside of it, how can the "few" charges in a conductor balance the field? From my point of view, it's the conductor surface which stops their movement. I'm obviously missing something fundamental here. Thanks in advance.

Answer

The common phrase is that metals have "electrons to burn", in other words, a mind-bogglingly large amount of free electrons, as @dmckee pointed out. Now, the electric field near a sheet of charge density $\sigma$ is $E = \frac{\sigma}{2\epsilon_0}$. Now $\epsilon_0 \approx 8.85 \times 10^{-12} F/m$ and Faraday's constant gives us an order of magnitude for the number of Coulombs in a mole of electrons to be $\approx 9.6\times 10^4$, so a Coulomb of charge per square meter of metal will yield a surface field of $\approx 10^{12+5-1} \sim 10^{16} V/m$ if I did my calculations right.

How "powerful" an electric field did you have in mind? Because as @Marshall pointed out, we are already beyond the breakdown fieldstrength here, so the number of electrons is not the limiting factor. To give you an idea, at $10^{18}V/m$, the electric field in a vacuum is strong enough to create electron-positron pairs (Schwinger mechanism) causing it to conduct. That's only about two orders of magnitude away.