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 $P_0$ 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?
On
Note that, according to your equation,
$$\frac{\mathrm{d}P}{\mathrm{d}t} = rP,$$ and $P(t=0) = P_0$. If $t$ indicates time and $P$ population, what is $r$? What does this equation tell us? How do you integrate it in order to get your solution?
On
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.
On
We define $e = \displaystyle \lim_{n\to \infty} \left(1+\dfrac{1}{n}\right)^n$. The reason it shows up, why we have this definition, is the following:
Say a population, $P_0$ doubles in a "unit" of time, $t'$ (we set this to $1$ for convenience right now). Then the population at a unit of time $t'$ is $P_0 *\left(1+\frac{1}{1}\right)$. Say you want to chop this up more say it up more, like twice. This means that at half a unit of time, you get half of the population to grow, and at full you get another half. The population is then $P_0 \left(1+\frac{1}{2}\right)^2$. In general, if you want to have your population growing $n$ times, the population in a unit of time is $\left(1+\frac{1}{n}\right)^n$
However, populations are ALWAYS growing, so you take the limit for continuously compounded population growth, and you get $e$.
Furthermore we can show that $e^{rt}=\lim\limits_{n\to\infty} \left(1+\frac{rt}{n}\right)^n$.
Now your question is, why can't we have a different base? Well the answer is that the $r$ takes care of it. Lets say you have a model:
$$P(t)=P_0 e^{rt}$$
and you want it in with a base $2$. Well simply note that $e^{rt}=(e^r)^t$. So if you want it in base $2$, let $r'=\log_2(e^{r})$. Then:
$$P(t)=P_0 e^{rt}=P_0 2^{\log_2(e^{r})t}=P_0 2^{r' t}$$
On
"e" has come in by solving a differential equation for continuous growth.
It is not necessarily there in a population growth equation, which could be of the form of
P = $P_0\cdot{(1+r)}^t$ or the even more general equation y = $a\cdot b^x$
The first is is of the form of compound interest, and the second uses growth factor "b" rather than growth rate (decimals) "r"
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 $e$.