Question: Let $p$ be a prime number and $$x\equiv a_1\pmod{p^{\alpha_1}} \\x\equiv a_2\pmod{p^{\alpha_2}} \\...\\x\equiv a_r\pmod{p^{\alpha_r}}$$ where $1\le\alpha_1\le\alpha_2...\le\alpha_r$. Show that the system has a simultaneous solution if and only if $a_i\equiv a_r\pmod{p^{\alpha_i}}$ for $i=1,2,3...,r$.
Lemma: $$x\equiv a_1\pmod{m_1}\\x\equiv a_2\pmod{m_2}$$ admits a solution if and only if $a_1\equiv a_2 \pmod{(m_1,m_2)}$.
Proof of lemma: Let $g=(m_1,m_2)$ and $m_i=m_i'g$ the system then is equivalent to solving $$x\equiv a_1\pmod{m_1'}\\x\equiv a_2\pmod{m_2'}\\x\equiv a_1\equiv a_2 \pmod{g}$$. By chinese remainder theorem the first two are solvable and from given condition the third one holds and so the system is solvable. The converse in which we assume the system has a solution and then we derive the third one are easy to show.
Solution (to question):We assume that $a_i\equiv a_r\pmod{p^{\alpha_i}}$ holds and then go on to show that the system has a solution. We proceed by induction on $r$. $r=2$ is given by lemma. I am having difficulties in the inductive step. Can I assume that a system of $r$ equations admits a solution, then introduce a new equation and carry on? I am a bit confused over here.
Please provide hints or any other methods of approach. Am I going correct?