Intermediate Radical Extension Paradox Suppose $K\subseteq L\subseteq M$ is a tower of fields.
If $M$ is a radical extension of $K$, is $L$ a radical extension of $K$?

My attempt (resulting in a paradox): 
I don't know much about radical extensions other than the definition. I tried cooking up this counter-example:
$K=\mathbb{Q}$
$L=\mathbb{Q}(\sqrt{1+\sqrt{2}})$
$M=\mathbb{Q}(\sqrt{2},\sqrt{1+\sqrt{2}})$
$M/K$ is radical since $\sqrt{2}^2$ lies in $\mathbb{Q}$, while $\sqrt{1+\sqrt 2}^2$ lies in $\mathbb{Q}(\sqrt{2})$.
$L/K$ is not radical since any power of $\sqrt{1+\sqrt 2}$ does not lie in $\mathbb{Q}$.
This seems to result in a paradox since $L=M$?
The definition of radical extension I am using is: An extension field $F$ of a field $K$ is a radical extension of $K$ if $F=K(u_1,\dots,u_n)$, some power of $u_1$ lies in $K$ and for each $i\geq 2$, some power of $u_i$ lies in $K(u_1,\dots,u_{i-1})$.

Thanks for any help.
Main question: Is the statement true or false and why?
Secondary question: Why is there a "paradox"? Radical extension depends on the generators used?
 A: We have $L=\mathbb{Q}(\sqrt{1+\sqrt{2}})$ and $M=L(\sqrt{2})$, so it should be $\sqrt{1+\sqrt{2}}^a\in \mathbb{Q}$ for some $a\in \mathbb{N}$, and $\sqrt{2}^b\in L$, for sombe $b\in \mathbb{N}$. But, there is no such $a$, hence $M/K=L/K$ isn't radical and there is no paradox.
A: As mentioned above:  "An extension field F of a field K is a radical extension of K if F=K($u_1,…,u_n$), some power of $u_1$ lies in K and for each i≥2, some power of $u_i$ lies in K($u_1$,…,u$_i$$_-$$_1$)".
All what the definition is saying is that if we can find a sequence of the u's which satisfies the conditions [namely: some power of $u_1$ lies in K and for each i≥2, some power of $u_i$ lies in K($u_1$,…,u$_i$$_-$$_1$)], then K/F is a radical extension.
If the u's are permutted then the extension field remains the same; that is K($u_1$,$u_2$) is the same as K($u_2$,$u_1$)
, for example.
However, the definition is NOT requiring that if K/F is to be radical then the conditions must be satisfied for all permutations of the u's.
So, to determine if $M = \mathbb{Q}(\sqrt{2},\sqrt{1+\sqrt{2}}) = \mathbb{Q}(\sqrt{1+\sqrt{2}},\sqrt{2})$ is radical over Q, take $u_1 = \sqrt{2}$ and $u_2 = \sqrt{1+\sqrt{2}}$ and you will end up with the "radical series" $Q\subseteq \mathbb{Q}(\sqrt{2})\subseteq\mathbb{Q}(\sqrt{2},\sqrt{1+\sqrt{2}})$; this is sufficient proof that M/$\mathbb{Q}$ is radical.
It is not necessary for "$Q\subseteq\mathbb{Q}(\sqrt{1+\sqrt{2}},)\subseteq \mathbb{Q}(\sqrt{1+\sqrt{2}},\sqrt{2})$" to be a radical series in order for M/$\mathbb{Q}$ to be radical.
[BTW: as already mentioned in the post above: $\mathbb{Q}(\sqrt{1+\sqrt{2}}) = \mathbb{Q}(\sqrt{1+\sqrt{2}},\sqrt{2})$]
