Fermat vs Euler

Fermat's little theorem states that for prime $p$ , we have $a^{p-1} \equiv 1\ (\mathrm{mod}\;p)$ $(p\not\mid a).$ Here is a proof:

1. Consider $1,2,\ldots,p-1$ and $a,2a,\ldots,(p-1)a.$
2. They are permutations of each other under $\mathrm{mod}\;p.$
3. $1\cdot 2\cdots (p-1)\equiv a\cdot 2a \cdots (p-1)a \;(\mathrm{mod}\;p).$
4. Cancellation: $1\equiv a^{p-1}\;(\mathrm{mod}\;p).$

However, when we are talking about a composite number $n$ , we have $a^{\phi(n)} \equiv 1\ (\mathrm{mod}\;n)$ for coprime integers $a$ and $n$ instead, from Euler's theorem .

If I use the above proof flow for the Euler's theorem, in which step do I first make a mistake?