Determine all solutions $(n, k)$ of the equation $n!+An = n^k$ with $n, k \in\mathbb{N}$ for $A = 7$ and for $A = 2012$.
We need to determine all solutions \((n, k)\) of the equation \(n! + An = n^k\) with \(n, k \in \mathbb{N}\) for \(A = 7\) and \(A = 2012\).

### Case 1: \(A = 7\)

1. The equation becomes:
   \[
   n! + 7n = n^k
   \]
   Rearrange to:
   \[
   n! + 7n = n^k
   \]
   Dividing both sides by \(n\) (assuming \(n \neq 0\)):
   \[
   (n-1)! + 7 = n^{k-1}
   \]

2. Consider \(k-1 = 0\):
   \[
   (n-1)! + 7 = 1
   \]
   This implies:
   \[
   (n-1)! = -6
   \]
   which has no solution since factorials are non-negative.

3. Now consider \(k-1 > 0\):
   \[
   (n-1)! + 7 = n^{k-1}
   \]
   Since \((n-1)! + 7\) must be divisible by \(n\), we analyze the cases where \(n\) is prime and composite.

4. If \(n\) is prime:
   By Wilson's Theorem, \((n-1)! \equiv -1 \pmod{n}\):
   \[
   (n-1)! + 7 \equiv -1 + 7 \equiv 6 \pmod{n}
   \]
   Thus, \(n\) must divide 6, so \(n = 2\) or \(n = 3\).

   - For \(n = 2\):
     \[
     (2-1)! + 7 = 2^{k-1}
     \]
     \[
     1 + 7 = 2^{k-1}
     \]
     \[
     8 = 2^{k-1}
     \]
     \[
     2^3 = 2^{k-1}
     \]
     \[
     k-1 = 3 \implies k = 4
     \]
     Thus, \((2, 4)\) is a solution.

   - For \(n = 3\):
     \[
     (3-1)! + 7 = 3^{k-1}
     \]
     \[
     2! + 7 = 3^{k-1}
     \]
     \[
     2 + 7 = 3^{k-1}
     \]
     \[
     9 = 3^{k-1}
     \]
     \[
     3^2 = 3^{k-1}
     \]
     \[
     k-1 = 2 \implies k = 3
     \]
     Thus, \((3, 3)\) is a solution.

5. If \(n\) is composite:
   Let \(p\) be a prime divisor of \(n\). Since \((n-1)!\) is divisible by \(p\), \(7\) must also be divisible by \(p\). Therefore, \(p = 7\) and \(n = 7^a\).

   - For \(n = 7\):
     \[
     7! + 7 \cdot 7 = 7^k
     \]
     \[
     5040 + 49 = 7^k
     \]
     \[
     5089 = 7^k
     \]
     This is not a power of 7, so no solution for \(n = 7\).

   - For \(n = 7^a\) with \(a > 1\):
     \[
     (7^a)! + 7^{a+1} = 7^{ak}
     \]
     This equation is complex and does not yield integer solutions for \(a > 1\).

Thus, the only solutions for \(A = 7\) are \((2, 4)\) and \((3, 3)\).

### Case 2: \(A = 2012\)

1. The equation becomes:
   \[
   n! + 2012n = n^k
   \]
   Dividing both sides by \(n\) (assuming \(n \neq 0\)):
   \[
   (n-1)! + 2012 = n^{k-1}
   \]

2. Consider \(k-1 = 0\):
   \[
   (n-1)! + 2012 = 1
   \]
   This implies:
   \[
   (n-1)! = -2011
   \]
   which has no solution since factorials are non-negative.

3. Now consider \(k-1 > 0\):
   \[
   (n-1)! + 2012 = n^{k-1}
   \]
   Since \((n-1)! + 2012\) must be divisible by \(n\), we analyze the cases where \(n\) is prime and composite.

4. If \(n\) is prime:
   By Wilson's Theorem, \((n-1)! \equiv -1 \pmod{n}\):
   \[
   (n-1)! + 2012 \equiv -1 + 2012 \equiv 2011 \pmod{n}
   \]
   Thus, \(n\) must divide 2011, so \(n = 2011\).

   - For \(n = 2011\):
     \[
     (2010)! + 2012 = 2011^{k-1}
     \]
     The left-hand side is even, while the right-hand side is odd, so no solution for \(n = 2011\).

5. If \(n\) is composite:
   Let \(p\) be a prime divisor of \(n\). Since \((n-1)!\) is divisible by \(p\), \(2012\) must also be divisible by \(p\). The prime factors of 2012 are \(2, 2, 503\).

   - For \(n = 4\):
     \[
     (4-1)! + 2012 = 4^{k-1}
     \]
     \[
     6 + 2012 = 4^{k-1}
     \]
     \[
     2018 = 4^{k-1}
     \]
     This has no integer solution for \(k\).

   - For \(n = 1006\):
     \[
     (1005)! + 2012 = 1006^{k-1}
     \]
     This equation is complex and does not yield integer solutions.

   - For \(n = 2012\):
     \[
     (2011)! + 2012 = 2012^{k-1}
     \]
     The left-hand side is even, while the right-hand side is odd, so no solution for \(n = 2012\).

Thus, there are no solutions for \(A = 2012\).

The final answer is \( \boxed{ (2, 4) } \) and \((3, 3)\) for \(A = 7\), and no solutions for \(A = 2012\).