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To get the asymptotic, one can replace the factorial in the exact formula by the Stirling formula. Then the desired quantity can be estimated via $$ A_N=N^{\phi(N)}\cdot \prod_{d|N}(\sqrt{2\pi d}\cdot e^{-d})^{\mu(d)}=(N/e)^{\phi(N)} $$ in the sense $$ c\cdot A_N\le (\dots)\le C\cdot A_N. $$

To get the asymptotic, one can replace the factorial in the exact formula by the Stirling formula. Then the desired quantity can be estimated as follows: $$ c\cdot A_N\le (\dots)\le C\cdot A_N $$ with $$ A_N=N^{\phi(N)}\cdot \prod_{d|N}(\sqrt{2\pi d}\cdot e^{-d})^{\mu(N/d)}=(N/e)^{\phi(N)}. $$

To get the asymptotic, one can replace the factorial in the exact formula be the Stirling formula. Then the desired quantity can be estimated via $$ A_N=N^{\phi(N)}\cdot \prod_{d|N}(\sqrt{2\pi d}\cdot e^{-d})^{\mu(d)}=(N/e)^{\phi(N)} $$ in the sense $$ c\cdot A_N\le (\dots)\le C\cdot A_N. $$

To get the asymptotic, one can replace the factorial in the exact formula by the Stirling formula. Then the desired quantity can be estimated via $$ A_N=N^{\phi(N)}\cdot \prod_{d|N}(\sqrt{2\pi d}\cdot e^{-d})^{\mu(d)}=(N/e)^{\phi(N)} $$ in the sense $$ c\cdot A_N\le (\dots)\le C\cdot A_N. $$

To get the asymptotic, one can replace the factorial in the exact formula be the Stirling formula. Then the desired quantity can be estimated via $$ A_N=N^{\phi(N)}\cdot\prod_{d|N}\left(\frac{\sqrt{2\pi d}}{e^d}}\right)^{\mu(d)}=\left(\frac{N}{e}\right)^{\phi(N)}} $$ in the sense $$ c\cdot A_N\le (...)\le C\cdot A_N $$

To get the asymptotic, one can replace the factorial in the exact formula be the Stirling formula. Then the desired quantity can be estimated via $$ A_N=N^{\phi(N)}\cdot \prod_{d|N}(\sqrt{2\pi d}\cdot e^{-d})^{\mu(d)}=(N/e)^{\phi(N)} $$ in the sense $$ c\cdot A_N\le (\dots)\le C\cdot A_N. $$