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It is given that $A= 9967^{9967^{9968}} = x^x$ . Since $9967$ is a prime, then $x$ must be of the form $x=9967^a$ and then we jhave:

$\Rightarrow x^x = 9967^{a\dot{}9967^a} = 9967^{9967\dot{}9967^ {9967}} = 9967^{9967^{9968}}$

$\Rightarrow x = 9967^{9967}$

The factors of $x$ are $\{ 9967^0, 9967^1, 9967^2,...9967^{9967}\}$ .

Therefore, the number of factors of $x$ is $\boxed{9968}$ .

$\huge {9967^{9967^{9968}}}$

can be written as

$\huge {(9967^{9967})^{9967^{9967}}}$

Therefore $x= 9967^{9967}$

As $9967$ is a prime number, the total number of factors is $\boxed {9968}$

This solution is by @Kalash Verma

It is just the simple observation that $a^{a^{a+1}}=a^{a^a\cdot a}=(a^a)^{(a^a)}$