Why do we use base $e$ in population growth questions? I know that we need base e to differentiate but I don't see what makes this formula work.

$$ P = P_0 e^{rt} $$

where the 0 refers to initial population, $r$ the rate, and $t$ the time. 

Changing the base changes the curve, so why does base e work? I mean $r$ and $t$ are pretty straightforward numbers so there's no fancy constants (other than $e$). Why is it not base $2$ or something else?
Two issues: (1) Why is base $e$ used?; and (2) "Changing the base changes the curve." is wrong if you do things right.

Suppose the population doubles every $30$ years.  Then what is the population after $180$ years?

Notice that $180/30=6$, i.e. $6$ is the number of $30$-year periods and thus the number of doublings.  So the population will be $P_0\cdot2^6 = P_0\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2 = 64 P_0$.

What is the population after $t$ years?  The number of $30$-year periods is $t/30$, so that's the number of doublings, and it is $P_0 \cdot 2^{t/30}$.

Notice that base $e$ is nowhere used above.

But what is the instantaneous rate of change at time $t=0$?  It is
$$
\left.\frac{dP}{dt}\right|_{t=0} = P_0 \frac d {dt} 2^{t/30} = P_0 \cdot 2^{t/30} (\log_e 2)\cdot 1 {30}.
$$
If you're not dealing with instantaneous rates of change, then you don't need $e$.

Suppose now we're told the initial population is $P_0$ and it's growing at $P_0/40$ per year.  How long will it take to double?  It follows that
$$
P= P_0 e^{t/40}. \tag 1
$$
If $t$ is the doubling time, then $e^{t/40}=2$, so $\dfrac t {40} =\log_e 2$ and the doubling time is $t=40\log_e 2 = 27.725887\ldots\text{ years}$.  Again, we need $e$ only because instantaneous rates of change are involved.

Notice that $\dfrac d {dt} 8^t = (8^t\cdot\text{constant})$.  If the base had been $6$ rather than $8$, the constant would be different.  Only when the base is $e$ is the constant $1$.  That is what is "natural" about $e$.

Above we found that the doubling time is $27.725887\ldots$ years, and the first argument above shows that
$$
P = P_0 \cdot 2^{t/27.725887\ldots}. \tag 2
$$

Is $(1)$ different from $(2)$?  No.  They're the same.  Changing the base does not change the curve if $t$ is multiplied by the constant appropriate to the base in each case.

I've seen students write things like $P = P_0 e^{-(\ln 8) t}$ and then fail to do the simplification that says this is $P = P_0\left(\frac 1 8 \right)^t$.  If $t=2$, then you have $P=P_0\cdot\frac 1 {64}$, but sometimes students say $\ln 8 = 2.079$ and $e^{-(2.079)\cdot 2} = 0.015638804272$, and fail to notice that that is close to $1/64$, and think that by adding lots of digits they're making it very accurate.  Those later digits are garbage. Notice that $1/64 = 0.015625$ and compare that with that previous number.

$\displaystyle P = P_0 e^{-(\ln 8) t}$ is the same as $\displaystyle P = P_0\left(\frac 1 8 \right)^t$, so changing the base doesn't change the curve if things are done correctly.
Note that $e = \displaystyle \lim_{n\to \infty} \left(1+\dfrac{1}{n}\right)^n$, and this is the continuous growth or decay problem, then you take the limit as $n$ to infinity and get $e$.