Calculate the following integral

This is an alternative approach.

Let $X_i$ ($i=1,\cdots , n$) be independent uniform random variable in $[0,1]$.

What is the PDF of $M=\max (X_1, \cdots, X_n)$?

Then what is $\mathbf{E}[M]$?

One can see by induction that:

$$\int_0^1 x_1 dx_1 \int_0^{x_2 } dx_2 \cdots = \int_0^1 x_1 \frac{x_{n - (n-1)}^{n-1}}{(n-1)!} dx_1 =\frac{1}{(n+1) \times (n-1)!}$$

Therefore, the original integral is:

$$n! \times \frac1{(n+1) \times (n-1)!} = \frac{n}{n+1}$$

Note that $\max(x_1,x_2,\dots,x_n)=\max(x_1,\max(x_2,\dots,x_n))$. In addition, note that

$$\begin{align} \int_0^1\int_0^1\max(x,y)\,dx\,dy&=\int_0^1\left(\int_0^y y\,dx+\int_y^1 x\,dx\right)\,dy\\\\ &=\int_0^1 \left(\frac12+\frac12 y^2\right)\,dy \end{align}$$

Now, we can write

$$\begin{align} \int_0^1 \max(x_1,x_2,\dots,x_n)\,dx_1&=\int_0^1 \max(x_1,\max(x_2,\dots,x_n))\,dx_1\\\\ &=\int_0^{\max(x_2,\dots,x_n)}\max(x_2,\dots,x_n)\,dx_1+\int_{\max(x_2,\dots,x_n)}^1x_1\,dx_1\\\\ &=\frac12 +\frac12\left(\max(x_2,\dots,x_n)\right)^2 \end{align}$$

Then, observe that

$$\begin{align} \frac{1}{k}\int_0^1 \left(\max(x_k,\dots ,x_n)\right)^{k}\,dx_k&=\int_0^{\max(x_{k+1},\dots ,x_n)}\left(\max(x_{k+1},\dots ,x_n)\right)^k\,dx_k+\int_{\max(x_{k+1},\dots ,x_n)}^1 x_k^k\,dx_k\\\\ &=\frac{1}{k}\left(\frac{k}{k+1}\left(\max(x_{k+1},\dots,x_n)\right)^{k+1}+\frac{1}{k+1}\right)\\\\ &=\frac{1}{k(k+1)}+\frac{1}{k+1}\left(\max(x_{k+1},\dots,x_n)\right)^{k+1} \end{align}$$

Proceeding inductively, we find that

$$\begin{align} \int_0^1\cdots \int_0^1 \max(x_1,x_2,\dots,x_n)\,dx_1\cdots dx_n&=\frac1{(2)(1)}+\frac{1}{(3)(2)}+\frac{1}{(4)(3)}+\cdots +\frac{1}{(n+1)(n)}\\\\ &=\sum_{k=1}^n \frac{1}{k(k+1)}\\\\ &=\sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right)\\\\ &=\frac{n}{n+1} \end{align}$$