There is no expansive homeomorphism on $S^1$

No, your proof does not work at all, starting with the claim that the sequence $(g^n)$ converges to $1$.

The fact that there are no expansive self-homeomorphisms of the circle is nonelementary and was first proven in a combination of work by Bryant on one hand and by Jakobsen and Utz on the other:

  1. Bryant proved in his 1954 PhD thesis, published in two installments:

Bryant, B. F., Expansive self-homeomorphisms of a compact metric space, Am. Math. Mon. 69, 386-391 (1962). ZBL0107.16502.

Bryant, B. F., On expansive homeomorphisms, Pac. J. Math. 10, 1163-1167 (1960). ZBL0101.15504.

that there are no expansive self-homeomorphisms of $[0,1]$.

  1. In the follow-up work (which appeared earlier)

Jakobsen, J. F.; Utz, W. R., The nonexistence of expansive homeomorphisms on a closed 2-cell, Pac. J. Math. 10, 1319-1321 (1960). ZBL0144.22302.

Jakobsen and Utz used Bryant's result to prove nonexistence of expansive self-homeomorphisms of the circle (this reduction is easier than Bryant's work).

Lastly: Bryant's work was generalized later on by various authors, for instance, in

Williams, R. K., A note on expansive mappings, Proc. Am. Math. Soc. 22, 145-147 (1969). ZBL0177.25604.

where Williams gives a new proof of nonexistence of expansive self-maps of the unit interval (without the assumption that the map is a homeomorphism).


A more personal proof (maybe the same as one reference given by Moishe)

If you consider a homeo of the interval, your proof is almost correct. Obviously if $f$ is expansive so is $f^2$. Therefore you may assume in this case $f$ is increasing. Every point is attracting to a fixed point (but $1$ is not necessarily attracting as you claimed it). This implies $f$ can not be expansive.

Now if $f$ is a homeomorphism of the circle with rational rotation number (equivalently $f$ has a periodic point) you are reduced to the case of interval maps. If the rotation number is irrational and $f$ has no wandering domain, it is conjugated to a rotation by Denjoy theory. But $f$ being assumed expansive it can not have wandering domains. Therefore we are reduced to rotations which are clearly not expansive.