Understand the problem

Find all pairs of positive integers $ (n,k)$ so that $ (n+1)^k-1=n!$.

Source of the problem
Singapore MO 2008
Number Theory
Difficulty Level
Suggested Book
An Excursion in Mathematics

Start with hints

Do you really need a hint? Try it first!

Note that n+1 has to be a prime, because any proper prime divisor of n+1 would divide n! too.
Say n+1=p. Note that, for large enough p, both \frac{p-1}{2} and 2 are factors of (p-1)!. Thus, the factor (p-1) occurs twice in the expansion of (p-1)!.
Prove that, if (p-1)^2|p^k-1 then p-1|k.
Hint 3 implies that p-1\le k. Hence p^{p-1}-1\le p^k-1=(p-1)!. This is obviously false. Hence, p has to be small enough to avoid this situation. It is avoided precisely when p\in\{2,3,5\}. This corresponds to n\in\{1,2,4\}. The equations to be solved are 2^k=2, 3^k=3 and 5^k=25. Hence, the solutions are \{(1,1),(2,1),(4,2)\}.

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