Prove the expression for acceleration due to gravity at a height h above the Earth’s surface

What you are trying to show

Gravity gets weaker as you move away from Earth because the same mass is spread over a larger sphere. So we start from Newton’s law of gravitation and rewrite it in terms of the usual surface value $g$.

Start from Newton’s gravitational force

For a mass $m$ at distance $r$ from Earth’s center: $$F=\frac{GMm}{r^2}$$ where $G$ is the gravitational constant and $M$ is Earth’s mass.

Convert force into acceleration at height h

At height $h$ above the surface, the distance from the center is $$r=R+h.$$ The acceleration due to gravity there is force per unit mass: $$g_h=\frac{F}{m}=\frac{GM}{(R+h)^2}.$$

Rewrite the result using surface gravity g

At the surface ($r=R$), $$g=\frac{GM}{R^2}.$$ So $GM=gR^2$. Substitute into the expression for $g_h$: $$g_h=\frac{gR^2}{(R+h)^2}=g\left(\frac{R}{R+h}\right)^2.$$ That proves the required expression.

Useful small-height approximation (optional)

If $h\ll R$, then $$\left(\frac{R}{R+h}\right)^2=\left(\frac{1}{1+h/R}\right)^2\approx 1-\frac{2h}{R}.$$ So $$g_h\approx g\left(1-\frac{2h}{R}\right).$$