elementary number theory - Is this proof for $11^{10} equiv 1 pmod{100}$ correct?

Show that $11^{10} \equiv 1\pmod{100}$

here, I asked a question about this same problem and all the answers say that I prove this by binomial expansion.

But by doing binomial expansion to show this is the same as doing $11^{10}$ mod $\ 100$ in the calculator, no property is being showed to prove this.

But I think this proof might be right.

$\gcd(11, 100) = 1$

$11^{φ(100)} \equiv 1 \pmod{100}$

$11^{40} \equiv 1 \pmod{100}$ $\ ⇒$

$11^{10} *11^{10} *11^{10}* 11^{10}\equiv 1\pmod{100}$

by Looking everybody mod $\ 100$, I can say that $11^{10} \equiv 1\pmod{100}$

Answer

This doesn't work because the solutions to $x^4 \equiv 1 \pmod{100}$ are $$x \equiv 1, 7, 43, 49, 51, 57, 93, 99 \pmod{100}.$$ So from $$(11^{10})^4 \equiv 1 \pmod{100}$$ the best you can say is that $$11^{10} \equiv 1, 7, 43, 49, 51, 57, 93, \text{ or } 99 \pmod{100}.$$