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Why z^f = (h_1^f)^x \bmod \tilde{N} #1

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@lucky1024

Hi experts.

Thank you very much for disclosing the rationale for the attack. I'm reading the paper you published, but I came across something I don't understand.

In section 3:
For eliminating $h_2^\rho$, we let $h_2 = h_1^e \bmod \tilde{N}$ in which $e$ is a divisor of $\text{ord}(h_1)$. Now, raising $z$ to the power of $f = \frac{\text{ord}(h_1)}{e}$
But, why we have $z^f = (h_1^f)^x \bmod \tilde{N}$ ?

Can you provide some clues to derivation? Thank you very much.

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