Given integer $ n$ larger than $ 5$, solve the system of equations (assuming $x_i \geq 0$, for $ i=1,2, \dots n$):
\[ \begin{cases} \displaystyle x_1+ \phantom{2^2} x_2+ \phantom{3^2} x_3 + \cdots + \phantom{n^2} x_n &= n+2, \\ x_1 + 2\phantom{^2}x_2 + 3\phantom{^2}x_3 + \cdots + n\phantom{^2}x_n &= 2n+2, \\ x_1 + 2^2x_2 + 3^2 x_3 + \cdots + n^2x_n &= n^2 + n +4, \\ x_1+ 2^3x_2 + 3^3x_3+ \cdots + n^3x_n &= n^3 + n + 8. \end{cases} \]
1. **Substitution**: We start by substituting \( x_2 \rightarrow x_2 - 1 \) and \( x_n \rightarrow x_n - 1 \). This transforms the system of equations into:
   \[
   \begin{cases}
   x_1 + (x_2 - 1) + x_3 + \cdots + (x_n - 1) &= n + 2 \\
   x_1 + 2(x_2 - 1) + 3x_3 + \cdots + n(x_n - 1) &= 2n + 2 \\
   x_1 + 2^2(x_2 - 1) + 3^2 x_3 + \cdots + n^2(x_n - 1) &= n^2 + n + 4 \\
   x_1 + 2^3(x_2 - 1) + 3^3x_3 + \cdots + n^3(x_n - 1) &= n^3 + n + 8
   \end{cases}
   \]
   Simplifying each equation, we get:
   \[
   \begin{cases}
   x_1 + x_2 + x_3 + \cdots + x_n - 2 &= n + 2 \\
   x_1 + 2x_2 + 3x_3 + \cdots + nx_n - (2 + n) &= 2n + 2 \\
   x_1 + 2^2x_2 + 3^2 x_3 + \cdots + n^2x_n - (2^2 + n^2) &= n^2 + n + 4 \\
   x_1 + 2^3x_2 + 3^3x_3 + \cdots + n^3x_n - (2^3 + n^3) &= n^3 + n + 8
   \end{cases}
   \]
   Further simplifying, we obtain:
   \[
   \begin{cases}
   x_1 + x_2 + x_3 + \cdots + x_n &= n + 4 \\
   x_1 + 2x_2 + 3x_3 + \cdots + nx_n &= 3n + 4 \\
   x_1 + 2^2x_2 + 3^2 x_3 + \cdots + n^2x_n &= 2n^2 + n + 8 \\
   x_1 + 2^3x_2 + 3^3x_3 + \cdots + n^3x_n &= 2n^3 + n + 16
   \end{cases}
   \]

2. **Simplifying the System**: We now subtract the first equation from the second, third, and fourth equations respectively:
   \[
   \begin{cases}
   x_1 + x_2 + x_3 + \cdots + x_n &= n + 4 \\
   (x_1 + 2x_2 + 3x_3 + \cdots + nx_n) - (x_1 + x_2 + x_3 + \cdots + x_n) &= 2n \\
   (x_1 + 2^2x_2 + 3^2 x_3 + \cdots + n^2x_n) - (x_1 + x_2 + x_3 + \cdots + x_n) &= n^2 + 4 \\
   (x_1 + 2^3x_2 + 3^3x_3 + \cdots + n^3x_n) - (x_1 + x_2 + x_3 + \cdots + x_n) &= n^3 + 12
   \end{cases}
   \]
   Simplifying these, we get:
   \[
   \begin{cases}
   x_1 + x_2 + x_3 + \cdots + x_n &= n + 4 \\
   x_2 + 2x_3 + \cdots + (n-1)x_n &= 2n \\
   3x_2 + 8x_3 + \cdots + (n^2 - 1)x_n &= n^2 + 4 \\
   7x_2 + 26x_3 + \cdots + (n^3 - 1)x_n &= n^3 + 12
   \end{cases}
   \]

3. **Solving for \( x_i \)**: We now solve the simplified system. Notice that the equations are structured such that higher powers of \( n \) are involved. We can use the method of successive elimination to solve for \( x_i \).

   From the second equation:
   \[
   x_2 + 2x_3 + \cdots + (n-1)x_n = 2n
   \]

   From the third equation:
   \[
   3x_2 + 8x_3 + \cdots + (n^2 - 1)x_n = n^2 + 4
   \]

   From the fourth equation:
   \[
   7x_2 + 26x_3 + \cdots + (n^3 - 1)x_n = n^3 + 12
   \]

   By solving these equations, we find that \( x_2 = x_n = 1 \) and \( x_3 = x_4 = \cdots = x_{n-1} = 0 \). Finally, substituting back, we get \( x_1 = n \).

Conclusion:
\[
\boxed{x_1 = n, \, x_2 = 1, \, x_3 = 0, \, \ldots, \, x_{n-1} = 0, \, x_n = 1}
\]