Simple Field extensions. Suppose $[L:K]=p$ $p$=prime .Then there exists $u$ such that $L=K(u)$

Proof.If  $[L:K]=p$ $p$=prime then for $M$ an  arbitrary field extension must hold that  $$[L:K]=[L:M][M:K]$$ if $[L:K]=prime$ then either $[L:M]=1$ or $[M:K]$=1 if the first one is  true then $L=M$ for arbitrary field extension so it is true for $L=K(u)$ . 

Now if the second one is true $M=K$ and i cant find argument that can prove that exists $u$ such that $L=K(u)$ only that for any extension i get K=M therefore K(u)=K. 

THat is my question on how to solve the second part
Take a element u in L which is not in K consider the field K(u) which contains K properly and using multiplicative properties of degree we can show that degree of K(u) over K is 1 or p since is not degree 1 hence it must be p and hence it is equal to L itself because it contained in L also has same degree.
Take a element u in L which is not in K consider the field K(u) which contains K properly and using multiplicative properties of degree we can show that degree of K(u) over K is 1 or p since is not degree 1 hence it must be p and hence it is equal to L itself because it contained in L also has same degree.