Positive integer solutions to $a^b-b!=a$ 
Conjecture: The only solutions to $a^b-b!=a$ such that $a,b\in\Bbb Z^+$ are $(a,b)=(2,2)$ and $(2,3)$.

Evidently, we have $a<b$, and extending the domain of $a,b$ to $\Bbb R^+$ reveals that asymptotically as $a\to+\infty$ either $b\sim1$ or $b\sim e(a-1)$ (using Stirling's approximation). However I don't see a way to continue further. Can the conjecture be proven?
 A: 
We are to solve the equation: $$a^b-a=b!$$ over positive integers.

It is clear that $a,b>1$. Assume that $a \geqslant b$. It then follows:
$$b^b-b \leqslant b!$$
and this clearly holds only for $b=2$, with equality, forcing $(a,b)=(2,2)$.
Now, let $a<b$ and $p$ be a prime factor of $a$ (such a prime exists as $a>1$). If $p<a$, then we have:
$$ap \mid b! \implies p \mid(a^{b-1}-1) \implies p \mid 1$$
which is clearly a contradiction. Thus, we must have $a=p$, which makes the equation:
$$p^b-p=b!$$
It is clear that as $p \mid b!$ and $p^2 \nmid b!$, we have $p<b<2p$. Then, we have:
$$p>\frac{b}{2} \implies \bigg(\frac{b}{2}\bigg)^b-\frac{b}{2}<b! \implies b<6$$
Thus, we have $a<b<6$, forcing $a<5$. If $a=2$, we have $2^b-2=b!$ forcing $(a,b)=(2,3)$. If $a=3$, then we have $3^b-3=b!$. It is simple to check that this has no solutions for $b<6$.
Thus, the only solutions are $(a,b)=(2,2),(2,3)$.
Note : There are many elementary ways to prove that:
$$\bigg(\frac{b}{2}\bigg)^b-\frac{b}{2}<b! \implies b<6$$
One approach is by pairing terms that add up to $b$ in the product $b!$ with slight modifications, and using $x(b-x)<b^2$. Being a little careful at places like $x=1,2$, we can show $b<6$. Another approach can be to use calculus to show the inequality required.
