How can you show by hand that $ e^{-1/e} < \ln(2) $? By chance I noticed that $e^{-1/e}$ and $\ln(2)$ both have decimal representation $0.69\!\ldots$, and it got me wondering how one could possibly determine which was larger without using a calculator. 
For example, how might Euler have determined that these numbers are different? 

Here's a method that only uses rational numbers as bounds, keeping to small denominators where possible. It is elementary, but finding sufficiently good bounds requires some experimentation or at least luck.
We instead establish the inequality $$-\log \log 2 < \frac{1}{e}$$ of positive numbers, which we can see to be equivalent by taking the logarithm of and then negating both sides. Our strategy for establishing this latter inequality (which is equivalent to yet another inequality suggested in the comments) is to use power series to produce an upper bound for the less hand side and a larger lower bound for the right-hand side. To estimate logarithms, we use the identity $$\log x = 2 \operatorname{artanh} \left(\frac{x - 1}{x + 1}\right) ,$$ which yields faster-converging series and so lets us use fewer terms for our estimate.
Firstly,
$$\log 2 = 2 \operatorname{artanh} \frac{1}{3} .$$
Then, since the power series $$\operatorname{artanh} u = u + \frac{1}{3} u^3 + \frac{1}{5} u^5 + \cdots $$ has nonnegative coefficients, for $0 < u < 1$ any truncation thereof is a lower bound for the series, and in particular 
$$\log 2 = 2 \operatorname{artanh} \frac{1}{3} > 2\left[\left(\frac{1}{3}\right) + \frac{1}{3} \left(\frac{1}{3}\right)^3 + \frac{1}{5} \left(\frac{1}{3}\right)^5\right] = \frac{842}{1215} .$$
We'll use the same power series to produce an upper bound for $-\log \log 2$, but since we're nominally computing by hand and want to avoid computing powers of a rational number with large numerator and denominator, we'll content ourselves with a weaker rational lower bound for which computing powers is faster: Cross-multiplying shows that
$$\log 2 > \frac{842}{1215} > \frac{9}{13}$$ and so
$$-\log \log 2 < -\log \frac{9}{13} = 2 \operatorname{artanh} \frac{2}{11} .$$
This time, we want an upper bound for $2 \operatorname{artanh} \frac{2}{11}$. When $0 < u < 1$ we have
$$\operatorname{artanh} u = \sum_{k = 0}^\infty \frac{1}{2 k + 1} u^{2 k + 1} < u + \frac{1}{3} u^3 \sum_{k = 0}^\infty u^{2k} = u + \frac{u^3}{3 (1 - u^2)},$$ and evaluating at $u = \frac{2}{11}$ gives
$$2 \operatorname{artanh} \frac{2}{11} < \frac{1420}{3861} < \frac{32}{87} .$$
On the other hand, the series for $\exp(-x)$ alternates, giving the estimate
$$\frac{1}{e} = \sum_{k=0}^\infty \frac{(-1)^k}{k!} > \sum_{k=0}^7 \frac{(-1)^k}{k!} = \frac{103}{280} .$$
Combining our bounds gives the desired inequality
$$-\log \log 2 < \frac{32}{87} < \frac{103}{280} < \frac{1}{e} .$$
