Why is power of a signal equal to square of that signal?

If the signal is represented as a voltage \$v(t)\$ or a current \$i(t)\$ and it is connected to a (1 ohm) resistor, the power dissipated in the resistor is proportional to \$v^2(t)\$ or \$i^2(t)\$.

Apart from that, defining power as a positive, increasing function of the signal amplitude has useful mathematical properties.

$$ P = \frac{V^2} R $$

If you're driving a constant resistance then the power is proportional to the square of the voltage.

You can rewrite the equation substituting, from Ohm's Law, \$ V = IR \$:

$$ P = \frac {V^2} R = V \frac V R = VI $$

and again ...

$$ P = VI = (IR)I = I^2R $$

It's a convention in signal processing theory to consider the signal being applied to a 1 Ω load, hence the expressions you were given.

My professor told: "It's a standard result..."

Your professor was strictly correct. It would have been more helpful if he or she had explained the 1 Ω convention.

If you are calculating power levels in a practical system you'll need to take account of the actual system impedance.