Let $p$ be an odd prime of the form $p=4n+1$. [list=a][*] Show that $n$ is a quadratic residue $\pmod{p}$. [*] Calculate the value $n^{n}$  $\pmod{p}$. [/list]
1. **Show that \( n \) is a quadratic residue \(\pmod{p}\):**

   Given \( p = 4n + 1 \), we need to show that \( n \) is a quadratic residue modulo \( p \).

   Since \( p \equiv 1 \pmod{4} \), we know that \( -1 \) is a quadratic residue modulo \( p \). This means there exists an integer \( i \) such that \( i^2 \equiv -1 \pmod{p} \).

   Consider \( 16n \equiv -4 \pmod{p} \). We can rewrite this as:
   \[
   16n \equiv (2i)^2 \pmod{p}
   \]
   This implies that \( 2i \) is a square modulo \( p \).

   We need to check the cases for \( p \equiv 1 \pmod{8} \) and \( p \equiv 5 \pmod{8} \):

   - If \( p \equiv 1 \pmod{8} \), then \( 2 \) is a quadratic residue modulo \( p \). Since \( i \) is also a quadratic residue (as \( i^{p-1} = (-1)^{\frac{p-1}{4}} = 1 \pmod{p} \)), \( 2i \) is a quadratic residue modulo \( p \).

   - If \( p \equiv 5 \pmod{8} \), then \( 2 \) is not a quadratic residue modulo \( p \). However, \( i \) is also not a quadratic residue (as \( i^{\frac{p-1}{2}} = (-1)^{\frac{p-1}{4}} = -1 \not\equiv 1 \pmod{p} \)). Therefore, \( 2i \) is a quadratic residue modulo \( p \).

   In both cases, \( 2i \) is a quadratic residue modulo \( p \), which means there exists an integer \( a \) such that:
   \[
   a^4 \equiv n \pmod{p}
   \]

   Thus, \( n \) is a quadratic residue modulo \( p \).

2. **Calculate the value \( n^n \pmod{p} \):**

   From the previous part, we know that \( n \equiv a^4 \pmod{p} \) for some integer \( a \). Therefore:
   \[
   n^n \equiv (a^4)^n \pmod{p}
   \]

   Since \( p = 4n + 1 \), we have:
   \[
   n = \frac{p-1}{4}
   \]

   Substituting this into the expression, we get:
   \[
   n^n \equiv (a^4)^{\frac{p-1}{4}} \pmod{p}
   \]

   Simplifying the exponent:
   \[
   (a^4)^{\frac{p-1}{4}} = a^{p-1}
   \]

   By Fermat's Little Theorem, \( a^{p-1} \equiv 1 \pmod{p} \). Therefore:
   \[
   n^n \equiv 1 \pmod{p}
   \]

The final answer is \( \boxed{ 1 } \).