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I was reading up on the Sharkovsky ordering and on the Wikipedia page, it mentions that it is not a well-ordering.

However, in the ordering itself as well as the "Period Three implies chaos", wouldn't 3 be the least element then? Because a 3-period would imply all other periods, but I haven't seen any mention that some other period would imply a 3-period. Am I understanding this right?

Any help is appreciated! Thanks!

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    The set of powers of two does not have a first element under that order. – 2017-02-19

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A well-ordering is much more than just a linear order with a least element - for $(L, \prec)$ to be a well-order, every nonempty subset of $L$ has to have a $\prec$-least element.

For example, the nonnegative rationals have a least element ($0$) but the set of positive rationals is a nonempty subset with no least element - so, the nonnegative rationals are not well-ordered.

And, in the Sharkovskii order, the powers of two form a descending sequence $$1\succ 2\succ 4\succ 8\succ . . .,$$ so the set $\{$powers of two$\}$ is nonempty but doesn't have a least element in the Sharkovskii sense.

EDIT: Note - to forestall possible confusion down the road - that some people define the Sharkovskii order in the opposite way; see e.g. here. This order, too, is not a well-order - consider the set of odd numbers . . .