I've worked long on the proof of the fundamental theorem of Arithmetic and there is only one tiny piece left I can't wrap my head around. Suppose that $$\prod_{i=1}^r p_i^{m_i} = \prod_{j=1}^s q_j^{n_j} $$ where of course all $p_i, q_j$ are primes and lets suppose they are 'ordered', meaning $p_i < p_{i+1}$ and $q_j< q_{j+1}$ and all the exponents be nonzero.
At the point where I say that without the loss of generality let $p_1 \mid q_i^{n_i}$ for some $ i \in \lbrace 1,2, \dots , s \rbrace$ it is clear it follows that $p_1 | q_i$ (using Euclid's lemma over and over again) and because $p_1$ and $q_i$ are both supposed to be primes it follows that: $$p_1 = q_i $$ Now I want to show that $i=1$ and I asked my tutor. He said that I could easily show that by contradiction, assuming $i >1$. So I've been stuck on this point for a while now.
My attempt:
Let $p_1 = q_i$ for some $1<i\leq s$ then I obtain: $$q_i^{m_1} p_2^{m_2} \dots p_r^{m_r} = q_1^{n_1} q_2^{n_2} \dots q_i^{n_i} \dots q_s^{n_s} $$ So $q_i$ divides the left and the right side, so it divides some $q_j$ for $j \in \lbrace 1 , \dots , s \rbrace$ thus $q_i=q_j$ but this can only be if $i=j$ for all $j$ because the prime numbers listed on the right hand side are 'ordered' (from small to big). But if $i=j$ for all $j \in \lbrace 1 , \dots , s \rbrace$ then especially $i=j=1$ but $i>1$, is this the contradiction I was looking for and is my argumentation correct?
Sorry if this question is blunt, but I have been staring at equations today for too long and number theory is not my forte.