7. All positive integer pairs $(a, b)$ that satisfy the condition $a^{a}-b^{b}=a^{a-1} \cdot b^{b-1}+a+b$ are
7. $\left\{\begin{array}{l}a=3 \\ b=2\end{array}\right.$ Detailed explanation: $a>b \geqslant 1$, from the conditions we get $a^{a}>a^{a-1} \cdot b^{b-1} \Rightarrow a>b^{b-1}$, thus we have $a \geqslant b^{b-1}+1$, so $a \cdot b \geqslant b^{b}+b$, which means $b^{b} \leqslant a b-b$, and from the given conditions and the above results we get
$a^{a-1} \cdot b^{b-1}+a+b=a^{a}-b^{b} \geqslant a^{a}-(a b-b)=a^{a}-a b+b$,
$\therefore a+a b \geqslant a^{a-1}\left(a-b^{b-1}\right)$, thus $a^{2}=a+a(a-1) \geqslant a+a b \geqslant a^{a-1}\left(a-b^{b-1}\right) \geqslant a^{a-1}$,
which means $1 \geqslant a^{a-3}$, so $2 \leqslant a \leqslant 3$, it is easy to see: when $a=2$, we must have $b=1$, but this does not meet the problem's conditions; when $a=3$, $b^{b}<3$, so $b=2$ (since $b=1$ does not meet the problem's conditions), thus the positive integer solution to this problem is $\left\{\begin{array}{l}a=3 \\ b=2\end{array}\right.$.