Well known problem:
Given an infinite grid of 1 ohm risistors, what is the resistance
across a single diagonal?

Well known answer:
2/pi ohms.

Question:
Where can I find a full solution to this problem?

Lets call the resistance of the grid Req. Since the grid is infinite
you could "break off" a segment, and the resistance of the remaining
grid will still be Req. The remaining segment is in parallel with one
of the 1 ohm resistors that has been "broken off."  The equation for
reisistors R1 and R2 in parallel is: Rp = R1*R2/(R1 + R2) . There is
still a one ohm resistor from the broken-off section that is in series
with Req.

The equivalent resistance Req  can be calculated using the equation:

Req = 1 + (Req * 1)/(1+Req)
Then
Req^2 - Req - 1 = 0
Use quadratic equation to solve

To clarify: the grid in question is infinite in 2 dimensions, not just
1.  The method mentioned by scodad works fine for an infinite "ladder"
of resistors, but since the answer has a pi in it, there is no point
in looking for a solution by such a simple algebraic method in the
case of an infinite (in 2 dimensions) grid.

By full solution, do you mean across more than one node?  If so, the answer is:
R(n,n) = 2/pi * Sum{k=1...n: 1/(2*k-1)}
R(1,1) = 2/pi
R(2,2) = 2/pi * (1 + 1/3)
R(3,3) = 2/pi * (1 + 1/3 + 1/5)

the derivation of this can be found among the following

http://mathquest.com/discuss/sci.math/a/m/277937/277940
 (with embedded links)

Other links...
http://hilbert.dartmouth.edu/~doyle/docs/net/net/net.html#thomasseninfinite
 (with embedded links)

Comments on a three dimensional grid...
http://mathforum.org/discuss/sci.math/m/193340/193357

A current link showing the integral solution for one 2D case

http://www.geocities.com/frooha/grid/node2.html