Good question.

Assume we have one cube of ice in a glass of water.  The ice displaces some of that water, raising the height of the water by an amount we will call $h$.

Archimedes principles states that the weight of water displaced will equal the upward buoyancy force provided by that water.  In this case, 

Weight of water displaced = $m_{waterdisplaced}g = \rho Vg = \rho Ahg$

Where $V$ is volume of water displaced, $\rho$ is density of water, $A$ is the surface area of the glass and $g$ is acceleration due to gravity.

Therefore the upward buoyancy force acting on the ice is $\rho Ahg$.

Now the downward weight of ice = $m_{ice}g$

Now because the ice is neither sinking nor floating, these must balance.  That is:

$\rho Ahg = m_{ice}g$

Therefore,

$h = \frac{m_{ice}}{\rho A}$

Now when the ice melts, this height difference due to buyoancy goes to 0.  But now an addition mass $m_{ice}$ of water has been added to the cup in the form of water.  Since mass is conserved, the mass of ice that has melted has been turned into an equivalent mass of water.

The volume of such water added to the cup is thus:

$V = \frac{m_{ice}}{\rho}$
and therefore,

$Ah = \frac{m_{ice}}{\rho}$

So,

$h = \frac{m_{ice}}{\rho A}$

That is, the height the water has increased due to the melted ice is exactly the same as the height increase due to buoyancy before the ice had melted.