Which is bigger

$\large A=2019^{\left(2019^{2019}\right)} \quad \text{or} \quad B=\left({2019^{2019}}\right)^{2019}?$

$B > A$
$A > B$
$A = B$

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$B=({2019^{2019}})^{2019}=2019^{2019 \cdot 2019}$

$2019^{2019} > 2019 \cdot 2019$ .

So $\boxed{A>B}$ .