How to show that $\frac{\pi}{5}\leq\int_0^1 x^x\,dx\leq\frac{\pi}{4}$

Changing variables $x\mapsto e^{-x}$ yields $$ \begin{align} \int_0^1(x\log(x))^n\,\mathrm{d}x &=\int_\infty^0(-xe^{-x})^n\,\mathrm{d}e^{-x}\\ &=(-1)^n\int_0^\infty x^ne^{-(n+1)x}\,\mathrm{d}x\\ &=\frac{(-1)^n}{(n+1)^{n+1}}\int_0^\infty x^ne^{-x}\,\mathrm{d}x\\ &=\frac{(-1)^nn!}{(n+1)^{n+1}}\tag{1} \end{align} $$ Pluging $(1)$ into $\displaystyle e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$ gives us $$ \int_0^1x^x\,\mathrm{d}x=\sum_{n=0}^\infty\frac{(-1)^n}{(n+1)^{n+1}}\tag{2} $$ As an alternating series with decreasing absolute values, we know that by using $(2)$, $$ \begin{align} \int_0^1x^x\,\mathrm{d}x &>1-\frac14\\ &=\frac34\\ &>\pi/5\tag{3} \end{align} $$ and $$ \begin{align} \int_0^1x^x\,\mathrm{d}x &<1-\frac14+\frac{1}{27}-\frac{1}{256}+\frac{1}{3125}\\ &=\frac{16922537}{21600000}\\ &<\pi/4\tag{4} \end{align} $$

It's already been mentioned in the comments that the minimum of the integrand (which is $(1/\mathrm e)^{1/\mathrm e}$, not $\mathrm e^{1/\mathrm e}$) is greater than $\pi/5$. However, proving that $(1/\mathrm e)^{1/\mathrm e}\gt\pi/5$ without a calculator would probably be rather tedious. A bound for which this would be slightly easier can be obtained by using the convexity of the exponential function:

$$ \begin{align} \int_0^1x^x\mathrm dx=\int_0^1\exp(x\log x)\,\mathrm dx\ge\exp\left(\int_0^1x\log x\,\mathrm dx\right)=\exp\left(-\frac14\right)\gt\frac\pi5\;.\end{align} $$

You still need to evaluate a couple of terms of some series whose error bounds you know in order to prove the last inequality, but it should be a bit easier than for $(1/\mathrm e)^{1/\mathrm e}$.