how to integrate $\int\underbrace{x^{x^{\cdot^{\cdot^x}}}}_ndx$ how to integrate 

$$\int\underbrace{x^{x^{\cdot^{\cdot^x}}}}_ndx$$ $\color{red}{\text{or how to calculate  this integral  when its bounded}}$
  $$\color{red}{\int_0^1\underbrace{x^{x^{\cdot^{\cdot^x}}}}_ndx}$$

Thanks in advance. 
$\color{green }{\text{my attempt}}$ :
its easy to integrate $\int x^xdx$ $$\int{x^xdx} = \int{e^{\log x^x}dx} = \int{\sum_{k=1}^{\infty}\frac{x^k\log^k x}{k!}}dx=  \sum_{k=0}^\infty \frac{1}{k!}\int x^k(\log x)^k\,dx \Rightarrow$$ substitute ${u = -\log x}$ then  $$ \int x^xdx=\sum_{k=0}^\infty \frac{(-1)^k}{k!}\int e^{u(k+1)}u^k\,du=\sum_{k=0}^\infty \frac{(-1)^k}{k!}\frac{1}{(k+1)^k}\int e^{u(k+1)}[(k+1)u]^k\,du.$$
Ii substitute  $t = (k+1)u$ and  $$\sum_{k=0}^\infty \frac{(-1)^k}{k!}\frac{1}{(k+1)^k}\int e^tt^k\,dt $$ if i put bound for this integral we have $$\int _0^1x^xdx=\sum_{k=0}^\infty \frac{(-1)^k}{k!}\frac{1}{(k+1)^k}\int_0^{\infty} e^tt^k\,dt =\sum_{k=0}^\infty \frac{(-1)^k}{(k+1)!}\frac{1}{(k+1)^k}\Gamma(k+1)=\sum_{k=0}^\infty \frac{(-1)^k}{(k+1)^{k+1}} = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^n}$$
$$\int_0^1\underbrace{x^{x^{\cdot^{\cdot^{x}}}}}_ndx=\int_0^1e^{\log\underbrace{x^{x^{\cdot^{\cdot^{x}}}}}_n}dx=\sum_{k=0}^\infty\frac{1}{k!}\int_0^1\biggl(\underbrace{x^{x^{\cdot^{\cdot^{x}}}}}_{n-1}\biggr)^k(\log x)^k~dx$$

MathWorld has already done for this.
Please see the formula (10) in http://mathworld.wolfram.com/PowerTower.html.
