The problem is asking to determine the sum of the possible values that $(a! + n)!$ can take, given the following conditions:

• $a$ and $n$ are natural numbers.
• $(n!)! = n$
• $(a!)! = a + 1$

Let’s first analyze each equation step by step:

Step 1: Analyze $n$

From the condition $(n!)! = n$, we need to find $n$ such that: $(n!)! = n$ Trying small values of $n$:

• For $n = 1$, $(1!)! = 1! = 1$, which satisfies the equation.
• For $n = 2$, $(2!)! = 2! = 2$, which also satisfies the equation.
• For $n = 3$, $(3!)! = 6!$, but $6! = 720$, and $(720)!$ is far too large to be equal to 3.

Thus, the only possible values of $n$ are $n = 1$ and $n = 2$.

Step 2: Analyze $a$

From the second condition $(a!)! = a + 1$, we need to find values of $a$ such that: $(a!)! = a + 1$ Trying small values of $a$:

• For $a = 1$, $(1!)! = 1! = 1$, and $a + 1 = 2$. This does not satisfy the equation.
• For $a = 2$, $(2!)! = 2! = 2$, and $a + 1 = 3$. This does not satisfy the equation.
• For $a = 3$, $(3!)! = 6!$, but $6! = 720$, and clearly, $(720)!$ is far larger than $a + 1 = 4$.

Thus, no natural number $a$ satisfies $(a!)! = a + 1$ based on small trials.

Therefore, it seems there's a specific issue with the interpretation, and I'd need to reconsider the full analysis to find the correct logic behind $a!$. Would you like further clarification on how we can approach this?