Without loss of generality, $a\le b$. If $a=1$, then $b!+1=b+1$ gives $b!=b$, which clearly implies $b\in\{1,2\}$. The resulting pairs $(1,1)$, $(2,1)$, $(1,2)$ indeed are solutions. From now on, let's assume $a,b\ge 2$. Because of AM-GM, for every integer $n\ge 2$, we have $$(n!)^{1/n}< \frac{1+2+\dots+n}{n}=\frac{n+1}2,$$ so $n!<\left(\frac{n+1}2\right)^n$ (equality cannot hold in the estimate). This shows that if $a=b$, then $a!=a^a$, while $a!<\left(\frac{a+1}2\right)^a\le a^a$ gives us a contradiction. From now on, assume $2\le a<b$. Now, for $b>a\ge 2$, $$a!+b!< \left(\frac{a+1}2\right)^a+\left(\frac{b+1}2\right)^b\le b^a+a^b,$$ if $a\ge \frac{b+1}2$. (Obviously, $\frac{a+1}2\le b$.) This leaves us the case $a\le \frac b2$. In the latter case, let $p$ be some prime divisor of $a$. Then since $2a$ is a factor of $\frac{b!}{a!}$, $p$ cannot divide $\frac{b!}{a!}+1$, and thus $$v_p(a!+b!)=v_p(a!)<\frac{a}{p-1}\le a,$$ by Legendre's formula. On the other hand, taking the equation modulo $p$ gives us $p\mid b^a$, so $p\mid b$. Hence, $p^a$ divides the RHS, and so $v_p(a!+b!)\ge a$. This gives a contradiction. Hence the solutions are $(1,1)$, $(2,1)$ and $(1,2)$.

$a=b\implies a!=a^a$ $\iff a=1$ (since $1\cdot 2\cdots a<a\cdot a\cdots a$ for $a\ge 2$). Wlog $a>b$. Two cases: 1) $\gcd(a,b)>1$. Let $p\mid \gcd(a,b)$, $b=pn$. Then $(pm)!+(pn)!=(pm)^{pn}+(pn)^{pm}$ with $m>n$. (See 1st page here for what $\upsilon_p(a)$ means). $\upsilon_p\left((pn)!\right)<\upsilon_p\left((pm)!\right)$. $\upsilon_p\left(\text{LHS}\right)=\upsilon_p\left((pn)!\right)=\lfloor\frac{pn}{p}\rfloor+\lfloor\frac{pn}{p^2}\rfloor+\cdots$ $<n+\frac{n}{p}+\cdots$ $=\frac{pn}{p-1}\le pn\le \upsilon_p\left(\text{RHS}\right)$ Contradiction. 2) $\gcd(a,b)=1$. Let $a=b+k, k>0, \gcd(b,k)=1$. $(b+k)!+b!=(b+k)^b+b^{b+k}$, so (Binomial theorem may help you see this) $b\mid k^b\iff b=1\implies (a,b)=(2,1)$. Answer: $\{(1,1),(1,2),(2,1)\}$ is the set of solutions.

Me We are given: \[ a!+b! = a^b+b^a \]using $\pmod{b}$ gives us that $b^a \equiv 0 \pmod{a}$ which gives us $p \mid b \implies p \mid a$. Now let $p \mid b$ be a prime. We by the properties of $\nu_p$ get: \[ \nu_p(a!+b!) \le \frac{b}{p-1} \le b \le \nu_p(a^b+b^a) \]which means $\nu_p(b!) = b \implies b$ is a prime or $1$. Putting it back yields the results stated above.

L_13832 For $a,b\leq 2$ we have $\boxed{(a,b)=(1,1),(2,1),(1,2)}$. WLOG $a>b>2$, so we have $b\mid a^b$, now take p such that $p\mid b$. The LHS can be simplified to $\left(\frac{a!}{b!}+1\right)b!$, also because $p\mid a$ we get $p\mid \frac{a!}{b!}$, these 2 conditions inturn give us that $p\mid b\implies p\mid a$. Due to all this simplification all we need to check is $\nu_p(b!)$ which is less than $\frac{b}{p-1}<b$. Looking at the RHS we have $\nu_p{(a^b+b^a)}\geq b$, so we have a contradiction.

The ideas are similar but the mode of execution is different