I am trying to prove that $$[k[u]:k[u^l]]=1$$ if $(l,[k[u]:k])=1$.
I do not know whether the above statement is true or false. Here $u$ is an algebraic number over the field $k$. But I was able to prove this in a very special case where $l$ is $2 $ and $[k[u]:k]$ is an odd integer. The method I used was $[k[u]:k[u^l]][k[u^l]:k]=[k[u]:k]$ and then I used the fact that $[k[u^l]:k] \geq n+1$. Now this sort of method is not generalising. If anyone can give any hints it would be great.
Well, I suppose that in your notation $k[u]$ is the smallest extension of $k$ wich contains $u$. In my notation this field is $k(u)$
Now, in this hypothesis, the assert is actually FALSE.
Let $k$ be $\mathbb{Q}$ and consider $u=\zeta_{15}$, where $\zeta_{15}$ is a primitive rooth of $x^{15}-1$.
We have $[\mathbb{Q}(\zeta_{15}):\mathbb{Q}]=\varphi(15)=8$, so we can take $l=3$ but $\zeta_{15}^3$ is a primitive 5-th rooth of 1, so a rooth of the polynomial $x^5-1$ and $$[\mathbb{Q}(\zeta_{15}^3):\mathbb{Q}]=\varphi(5)=4 < [\mathbb{Q}(\zeta_{15}):\mathbb{Q}]=8$$
and finally we obtain $[\mathbb{Q}(\zeta_{15}):\mathbb{Q}(\zeta_{15}^3)]=2$.