Although this has been answered a couple of times I would like to add the reasoning that I personally find most eye opening and is taken from Tom Lee's book "Planar Microwave Engineering" (chapter 2.3).
As indicated in the other responses, most people forget that Kirchoffs laws are just approximations that hold under certain conditions (the lumped regime) when quasi-static behavior is assumed. How does it come to these approximations?
Let's start with Maxwell's quations in free space:
$$
\nabla \mu_0 H=0 \qquad(1) \\
\nabla \epsilon_0 E = \rho \qquad(2) \\
\nabla\times H=J+\epsilon_0 \frac{\partial E}{\partial t} \qquad(3) \\
\nabla\times E=-\mu_0 \frac{\partial H}{\partial t} \qquad(4) \\
$$
Equation 1 states that there is no divergence in the magnetic field and hence no magnetic monopoles (mind my username! ;-) )
Equation 2 is Gauss' law and states that there are electric charges (monopoles). These are the sources of the divergence of the electric field.
Equation 3 is Ampere's law with Maxwells modification: It states that ordinary current as well as a time-varying electric field creates a magnetic field (and the latter one corresponds to the famous displacement current in a capacitor).
Equation 4 is Faradays law and states a changing magnetic field causes a change (a curl) in the electric field.
Equation 1-2 is not important for this discussion but Equation 3-4 answer where the wave behavior comes from (and since Maxwell's equations are most generic, they apply to all circuits incl DC): A change in E causes a chance in H which causes a change in E and so forth. Is is the coupling terms that produce wave behavior!
Now assume for a moment mu0 is zero. Then the electrical field is curl free and can be expressed as the gradient of a potential which also implies that the line integral around any closed path is zero:
$$
V = \oint E dl = 0
$$
Voila, this is just the field-theoretical expression of Kirchhoff's Voltage Law.
Similarly, setting epsilon0 to zero results in
$$
\nabla J = \nabla (\nabla \times H) = 0
$$
This means the divergense of J is zero which means no (net) current can build up at any node. This is nothing more than Kirchhoffs Current Law.
In reality epsilon0 and mu0 are of course not zero. However, they appear in the definition of the speed of light:
$$
c = \sqrt{\frac{1}{\mu_0 \epsilon_0}}
$$
With infinite speed of light, the coupling terms would vanish and there would be no wave behavior at all. However, when the physical dimensions of the system are small compared to the wavelengths then the finiteness of the speed of light is not noticeable (similarly as time dilation always exist but won't be noticable for low speeds and hence Newtons equations are an approximation of Einsteins relavivity theory).