$\mu_R(m)\geq \dim R$ where $\mu_R(m)=\operatorname{embdim}R$.

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Let ($R,m,k$) be a noetherian local ring. Show that $\mu_R(m)\geq \dim R$ where $\mu_R(m)=\operatorname{embdim}R$.

It results by Krull's Principal Ideal Theorem but I don't know how.

Principal Ideal Theorem: If $R$ is a noetherian ring and $I=(a_1,...,a_n)$ then $\mathrm{ht}(P)\leq n$ for any minimal prime ideal $P$ that contains $I$.