$x^{y}$ is $y^{x}$

The only distinct positive integers for which the given equation is true are 2 and 4: $2^4 = 4^2 = 16.$ Then $2^2\cdot 4^4 = \boxed{1024}$

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Proof that this is the only solution. Assume without loss of generality that $x > y \geq 1$ .

Let $a = \text{gcd}(x,y)$ . Then $u = x/a$ and $v = y/a$ are coprime. The equation $x^y = y^x$ yields $a^yu^y = a^xv^x \ \ \therefore\ \ u^y =a^{x - y}v^x.$ Since $u$ and $v$ are coprime, they have no common prime factors. On the other hand, any prime factor in $v$ must also be a prime factor in $u$ . This implies $v = 1;\ \ y = a;\ \ u = \sqrt[y]{a^{x-y}} = a^{x/y-1};\ \ x = a^{x/y} = a^u.$ We see that $u = a^{u - 1}$ . It is also clear that $a > 1$ and $u > 1$ .

Now set $b = a - 1 > 0$ and $w = u - 1 > 0$ ; by binomial expansion we write $a^{u-1} = (1 + b)^w = 1 + wb + \cdots.$ But this must be equal to $u = 1 + w$ ; this only works if $b = 1$ and $w = 1$ , i.e. $a = 2$ and $u = 2$ . This means of course that $x = 4,\ \ y = 2.$