Where's my mistake? This is partially an electrical engineering problem, but the actual issue apparently lays in my maths.
I have the equation $\frac{V_T}{I_0} \cdot e^{-V_D\div V_T} = 100$
$V_T$ an $I_0$ are given and I am solving for $V_D$.
These are my steps:
Divide both sides by $\frac{V_T}{I_0}$: 
$$e^{-V_D\div V_T} = 100 \div \frac{V_T}{I_0}$$
Simplify: 
$$e^{-V_D\div V_T} = \frac{100\cdot I_0}{V_T}$$
Find the natural log of both sides: 
$$-V_D\div V_T = \ln(\frac{100\cdot I_0}{V_T})$$
Multiply by $V_T$: 
$$-V_D = V_T \cdot \ln(\frac{100\cdot I_0}{V_T})$$
Multiply by $-1$:
$$V_D = -V_T \cdot \ln(\frac{100\cdot I_0}{V_T})$$
Now if I substitute in the numbers; $V_T \approx 0.026$, $I_0 \approx 8 \cdot 10^{-14}$
$$V_D = -0.026 \cdot \ln(\frac{8 \cdot 10^{-12}}{0.026})$$
Simplify a little more:
$$ V_D = \frac{26}{1000} \cdot -\ln(\frac{4}{13} \cdot 10^{-9})$$
and you finally end up with $V_D \approx 0.569449942$
There is an extra step to the problem as well: the question calls not for $V_D$, but for $V_I$ which is a source voltage and that can basically be determined by solving this:
$$V_I \cdot \frac{1}{40} = V_D$$
I.e. $V_I = 40V_D$; which makes $V_I \approx 22.77799768$.
However, this is off by quite a bit (the answer is apparently $1.5742888791$).
Official Solution:

We find $V_D$ to be $\approx$ 0.57.

(Woo! I did get that portion right.)

Since we know that $\frac{V_D}{i_D}$ is 100, we find $i_D$ to be 0.00026.

Background: $\frac{V_D}{i_D}$ is the resistance that I was originally solving. $V_D$ was the voltage across the element, and $i_D$ was the current through the element.
However, either I'm making some stupid mistake but if $i_D = \frac{V_D}{100}$ then how did they get 0.00026?
Completing the rest of the solution's method (quite convoluted in comparison to mine, checked mine though another method; $V_I$ does in fact equal $40V_D$), with the correct value, 0.0057, I arrived at exactly the same final value as before.
Would it be fair to say that it is likely that my logic is correct?

You are confusing resistance with incremental resistance, I think. The incremental resistance only matters for small signal analysis. The problem is to set the operating point so that the incremental resistance will be the required value. This involves computing $V_D, I_D$. However, you cannot use the incremental resistance to compute $I_D$ in terms of $V_D$.
Also, you are forgetting to account for the $3.9k\Omega$ series resistance.
You correctly computed $V_D = 0.57 V$ required so the incremental resistance is $100 \Omega$.
However, the current through the diode at $V_D$ is given by the diode equation, which you haven't included above. The diode equation is $I_D = I_0 (e^{V_D/V_T}-1)$. Plugging in your numbers gives $I_D = 260 \mu A$ (basically $\frac{V_T}{100}$, since $\frac{V_D}{V_T}$ is large).
The question was to figure the input voltage $V_I$ that will set the diode operating point at $V_D =0.57 V, I_D=260 \mu A$. In the first instance, there is a series resistance of $R_S = 3.9 k\Omega$, so to figure the required $V_I$ you need
$$ V_I = R_S I_D+V_D$$
Plugging in the numbers gives $V_I = 1.58V$ or so.
