With reference to fig.3a below,

The forward gain of the amplifier A is given by:
$$
A = \frac{I_o}{I_s}
$$
The feedback factor of the network $\beta$ is given by:
$$
\beta = \frac{I_f}{I_o}
$$

From the diagram, the input current to the amplifier A is the difference between the source current $I_{in}$ and the feedback current $I_f$:
$$
I_s = I_{in} - I_f
$$

Using the feedback factor definition, $I_f = \beta I_o$. Substitute this into the $I_s$ equation:
$$
I_s = I_{in} - \beta I_o
$$

From the forward gain definition, $I_s = \frac{I_o}{A}$. Substitute this into the equation from Step 3:
$$
\frac{I_o}{A} = I_{in} - \beta I_o
$$

Move the term with $I_o$ to one side:
$$
\frac{I_o}{A} + \beta I_o = I_{in}
$$
Factor out $I_o$:
$$
I_o \left( \frac{1}{A} + \beta \right) = I_{in}
$$
Combine the terms in the parenthesis:
$$
I_o \left( \frac{1 + A\beta}{A} \right) = I_{in}
$$
Finally, solve for the overall gain with feedback, $A_f$:
$$
A_f = \frac{I_o}{I_{in}} = \boxed{\frac{A}{1 + A\beta}}
$$