I have a question about properties of the multiplicative groups.

Let we have finite field of prime order $2^k$ -1.

It is clear that multiplicative group of such field has subgroup of order.

$$\frac{2^k-2}{k}$$.

How it is possible to find generator $g$ of this group (I think that this group is always cyclic).

For example for $k=5$ $g=6$ For example for $k=7$ $g=18$

I think that

$$k=\frac{2^k-2}{k}$$.

But can't prove it.