 Solution

Since $K=F(\lambda )$ and the extension is finite, the element $\lambda$ is algebraic. Its minimal polynomial is irreducible, and evaluation gives an isomorphism $F[X]/(m_{\lambda })\cong F[\lambda ]=F(\lambda )=K$. If its degree is $d$, the powers $1,\lambda ,\dots ,{\lambda }^{d-1}$ form a basis. Comparing with $[K:F]=n$ gives $\boxed{\mathrm{deg}{m}_{\lambda }=n}$.