Power of n equalling (n - 1)! + 1

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Theorem

There is exactly one solution to the equation in the integers:

$\paren {n - 1}! + 1 = n^k$

for $k > 1$, and that is:

$n = 5$
$k = 2$


Theorem

We have that:

\(\ds \paren {1 - 1}! + 1\) \(=\) \(\ds 0! + 1\)
\(\ds \) \(=\) \(\ds 1 + 1\) Factorial of Zero
\(\ds \) \(=\) \(\ds 2\) not a power of $1$
\(\ds \paren {2 - 1}! + 1\) \(=\) \(\ds 1! + 1\)
\(\ds \) \(=\) \(\ds 1 + 1\) Examples of Factorials
\(\ds \) \(=\) \(\ds 2\)
\(\ds \) \(=\) \(\ds 2^1\) $k = 1$
\(\ds \paren {3 - 1}! + 1\) \(=\) \(\ds 2! + 1\)
\(\ds \) \(=\) \(\ds 2 + 1\) Examples of Factorials
\(\ds \) \(=\) \(\ds 3\)
\(\ds \) \(=\) \(\ds 3^1\) $k = 1$
\(\ds \paren {4 - 1}! + 1\) \(=\) \(\ds 3! + 1\)
\(\ds \) \(=\) \(\ds 6 + 1\) Examples of Factorials
\(\ds \) \(=\) \(\ds 7\) not a power of $4$
\(\ds \paren {5 - 1}! + 1\) \(=\) \(\ds 4! + 1\)
\(\ds \) \(=\) \(\ds 24 + 1\) Examples of Factorials
\(\ds \) \(=\) \(\ds 25\)
\(\ds \) \(=\) \(\ds 5^2\)


Let $n > 5$.

Suppose $n$ is composite.

By Divisibility of n-1 Factorial by Composite n:

$n \divides \paren {n - 1}!$

and thus:

$n \nmid \paren {n - 1}! + 1$

showing that $\paren {n - 1}! + 1$ is not a power of $n$.


Suppose $n$ is prime.

Further suppose:

$\exists k \in \N: \paren {n - 1}! + 1 = n^k$

Then:

\(\ds \paren {n - 1}!\) \(=\) \(\ds n^k - 1\)
\(\ds \) \(=\) \(\ds \paren {n - 1} \sum_{i \mathop = 0}^{k - 1} n^i\) Sum of Geometric Sequence
\(\ds \leadsto \ \ \) \(\ds \paren {n - 2}!\) \(=\) \(\ds \sum_{i \mathop = 0}^{k - 1} n^i\)
\(\ds \) \(\equiv\) \(\ds \sum_{i \mathop = 0}^{k - 1} 1^i\) \(\ds \pmod {n - 1}\) Congruence of Powers
\(\ds \) \(\equiv\) \(\ds k\) \(\ds \pmod {n - 1}\)

By Divisibility of n-1 Factorial by Composite n, since $n - 1 \ne 4$ and is composite:

$n - 1 \divides \paren {n - 2}!$

thus $k$ must be a multiple of $n - 1$.


However:

\(\ds n^{n - 1}\) \(>\) \(\ds n \paren {n - 1} \paren {n - 2} \cdots \paren 2\)
\(\ds \) \(=\) \(\ds n!\)
\(\ds \) \(>\) \(\ds 2 \paren {n - 1}!\)
\(\ds \) \(>\) \(\ds \paren {n - 1}! + 1\)
\(\ds \) \(>\) \(\ds 1\)
\(\ds \) \(=\) \(\ds n^0\)

which shows that no multiple of $n - 1$ can satisfy our equation.

Hence there is no solution for $n > 5$.

$\blacksquare$


Historical Note

The fact that $25 = 4! + 1$ is the only solution to $\left({n - 1}\right)! + 1 = n^k$ was apparently proved by Joseph Liouville, but this requires corroboration.


Sources

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