Let $P_0 \subsetneq P_{1} \subsetneq \dotsb \subsetneq P_n$ be a chain of prime ideals in a Noetherian ring. Let $P_0 \subset P_{1} \subset \dotsb \subset P_n$ be a chain of prime ideals in a Noetherian ring.
Show that if $x \in P_n$ then there exists a chain of prime ideals $Q_{1} \subsetneq \dotsb \subsetneq Q_n = P_n$ such that $x \in Q_1$.

I'm lost on how to tackle this problem. Any help would be appreciated!
This is Lemma 2.5.13 in Qing Liu's book Algebraic Geometry and Arithmetic Curves. The proof here uses Krull's principal ideal theorem to employ induction on $n$. 

I find the proof quite non-trivial, definitely not something I would come up with. I don't know if there is an elementary way to prove it. As an aside, Liu then uses this lemma to deduce that Noetherian local rings have finite dimension (in fact the dimension is bounded by the number of generators needed for the unique maximal ideal) in Corollary 5.14.
This is Lemma 2.5.13 in Qing Liu's book Algebraic Geometry and Arithmetic Curves. The proof here uses Krull's principal ideal theorem to employ induction on $n$. 

I find the proof quite non-trivial, definitely not something I would come up with. I don't know if there is an elementary way to prove it. As an aside, Liu then uses this lemma to deduce that Noetherian local rings have finite dimension (in fact the dimension is bounded by the number of generators needed for the unique maximal ideal) in Corollary 5.14.