A number with an interesting property. $abcd=a^b c^d$ I am finding a 4-digit number $abcd$ (in base-10 representation) satisfying the following property.
$$abcd=a^b c^d$$
I have been running my mind around this problem since a long time, but got no success.
I wrote a python program for a number with this property, and got the answer to be $2592$ since 
$$2592=2^5\times 9^2$$
But I am looking for a purely mathematical way to solve this problem. 
Thanks!
 A: I will type up the answer provided in the published solution for posterity.
From the American Mathematical Monthly problem E$69$ (Vol. $41$, May $1934$, p. $322$)
(accessed via JSTOR)

Instead of a product of powers, $a^bc^d$, a printer accidentally prints the four-digit number, $abcd$. The value is however the same. Find the number and show that it is unique.

*

*Since $abcd = a^bc^d$, neither $a^b$ nor $c^d$ can have more than four digits. Hence the highest possible powers for any single digits in the problem are $9^4$, $8^4$, $7^4$, $6^5$, $5^5$, $4^6$, $3^8$, $2^9$, and $1^9$
$2.$

*Neither $a$ nor $c$ is zero, as that would make $abcd$ zero

*If $a=1$, or if $b=0$, then $abcd = c^d$. Similarly, if $c=1$ or $d=0$, then $abcd = a^b$. Examination of the expanded powers which have four digits shows that none meets these conditions. Therefore $a \not =1$, $b \not = 0$, $ c \not = 1$, and $d \not =0$.

*If one of $a^b$ and $c^d$ has four digits, the other is less than ten and so must be $n^1$, $2^2$, $2^3$, or $3^2$. But no product of an eligible four digits expanded power by any one of these numbers meets condition (1), so these possibilities are accordingly ruled out.

*Since $abcd$ cannot end in zero, and since neither $a$ nor $c$ is one, $2000 \lt abcd \lt 10,000$.

*A test of the possible products formable from the remaining eligible expanded factors by multiplying the units digits and noting that the resultant units digits is not the same as either exponent, eliminates the majority of possibilities. Complete multiplication disposes of all others except $2^5 \cdot 9^2 = 32 \cdot 81 = 2592$, which is therefore the unique solution.


