Prove that there are only finite numbers satisfying this equation:

Question

Coincidentally, I realized that my room number $(23)$ has the following property:

$$2^3+1=3^2$$

In order to find more numbers $n$ exhibiting this property, I wrote the following equation: $$(n-1)^n+1=n^{n-1}$$

Now, I realized that for $n>3$, where $n$ is an integer, the LHS becomes greater than the RHS and therefore if one shows that for $n>3$ the following inequality holds, it would prove that there are not infinite numbers that show this property.

Inequality: $(n-1)^n+1>n^{n-1}$

I am still a beginner in Inequalities and would, therefore, like to see as to how should one proceed further with this argument.

Answer 1

$(n-1)^n+1\gt n^{n-1}$ can be rewritten as

$$\left(1-{1\over n}\right)^n\gt{1\over n}-{1\over n^n}$$

Does the left hand side look familiar?

Answer 2

There is no other solution than that which is given; first observe that $n$ must be odd otherwise the $+1$ will not cancel, expanding as binomial; $[-1+n(n-1)/2-...+n^n +1]=n^{n-1}$ from here you notice that for the equation to hold for any odd $n>3$, then $n|n-1$, which is obviously false.