Euler totient function: if phi(m) =6 then why is m not divisible by any prime greater than 7?

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Came across this line in a book: If $\phi(m) = 6$, then $m$ is not divisible by any prime greater than $7$.

I can't seem to understand why this is the case; I know that $\phi(m)$ is defined in terms of its prime factorisation, but why is $7$ the biggest prime that can divide $\phi(m)$?

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We can show (or define it this way):

$$\phi(p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k})=p_1^{e_1-1}(p_1-1)p_2^{e_2-1}(p_2-1)\cdots p_k^{e_k-1}(p_k-1)$$

where $p_i$ are distict primes and $e_i\ge 1$ for all $i$.

Thus, if $p$ divides $n$, then $p-1\mid \phi(n)$.


So if we know $\phi(n)=6$, then for every prime $p$ that divides $n$, we know $p-1\mid \phi(n)=6$, so that $p$ can be no larger than $7$.