Unique Factorisation Theorem: Let $R$ be a Euclidean ring and $a\neq 0$ non-unit in $R.$ Suppose that $a =\pi_1\pi_2\cdots\pi_n=\pi_1'\pi_2'\cdots\pi_m'.$ where the $\pi_i$ and $\pi_j'$ are prime elements of $R.$ Then $n = m$ and each $\pi_i, 1\leq i\leq n$ is an associate of some $\pi_j',1\leq j\leq m$ and conversely each $\pi_k'$ is an associate of some $\pi_q.$
The proof given is as follows:
Look at the relation $a= \pi_1\pi_2\cdots\pi_n =\pi_1'\pi_2'\cdots\pi_m'.$ But $\pi_1|\pi_1\pi_2\cdots\pi_n$ hence $\pi_1|\pi_1'\pi_2'\cdots\pi_m'.$ So, $\pi_1$ must divide some $\pi_i'$; since $\pi_1$ and $\pi_i'$ are both prime elements of $R$ and $\pi_1|\pi_i'$ they must be associates and $\pi_i' = u_1\pi_1,$ where $u_1$ is a unit in $R.$ Thus $a=\pi_1\pi_2\cdots\pi_n =u_1\pi_1\pi_2'\cdots\pi_{i-1}'\pi_{i+1}'\cdots\pi_m',$ cancel off $\pi_1$ and we are left with $\pi_2\cdots\pi_n =u_1\pi_2'\cdots\pi_{i-1}'\pi_{i+1}'\cdots\pi_m'.$ Repeat the argument on this relation with $\pi_2.$ After $n$ steps, the left side becomes $1,$ the right side a product of a certain number of $\pi'$ (the excess of $m$ over $n$). This would force $n\leq m$ since the $\pi'$ are not units. Similarly, $m\leq n,$ so that $n = m.$ In the process we have also showed that every $\pi_i$ has some $\pi_i'$ as an associate and conversely.
However, I don't get the line,
Thus $a=\pi_1\pi_2\cdots\pi_n =u_1\pi_1\pi_2'\cdots\pi_{i-1}'\pi_{i+1}'\cdots\pi_m',$ cancel off $\pi_1$ and we are left with $\pi_2\cdots\pi_n =u_1\pi_2'\cdots\pi_{i-1}'\pi_{i+1}'\cdots\pi_m'.$
This is because, I feel that instead of the equality, $$a=\pi_1\pi_2\cdots\pi_n =u_1\pi_1\pi_2'\cdots\pi_{i-1}'\pi_{i+1}'\cdots\pi_m',$$ we must have, $$a=\pi_1\pi_2\cdots\pi_n =u_1\pi_1\pi_1'\pi_2'\cdots\pi_{i-1}'\pi_{i+1}'\cdots\pi_m'.$$ Where did the $\pi_1'$ disappear in that equality suddenly? So, after cancelling off $\pi_1$ we are left with the equality $\pi_2\cdots\pi_n =u_1\pi_1'\pi_2'\cdots\pi_{i-1}'\pi_{i+1}'\cdots\pi_m'.$
Secondly, I don't get the line,
After $n$ steps, the left side becomes $1,$ the right side a product of a certain number of $\pi'$ (the excess of $m$ over $n$).
We don't really know anything about $m,n\in\Bbb Z^+$, so we may have $m<n$ as well, i.e the left side might have more numbers than the right side and in that case, we won't have a $1$ in the left hand side. Also, the right side might not have any $\pi'$ if $m=n$ and in this case, the right side contains a product of a finite number of unit elements of $R.$ So, the conclusion made in the proof, i.e
This would force $n\leq m$ since the $\pi'$ are not units.
seems weird.
Therefore, the subsequent lines, i.e
Similarly, $m\leq n,$ so that $n = m.$
makes no sense to me. I need some clarifications regarding this proof.
The missing $\pi'_1$ just seems like a typo, that shouldn't be a problem.
As for the second part of your question, it could have been clearer but it's more an argument of the type "suppose by contradiction that $m > n$. (insert argument) hence we would have (contradiction), thus $m \leq n$. Conversely, suppose by contradiction that $n > m$ (...)".
They also got their directions wrong: they use the fact that the $\pi'$s are not units, thus you cannot have $1 = \pi'_{j_1} \dots \pi'_{j_p}$ for any subcollection of the $\pi'$s, but this implies that you need $n \geq m$ and not $n \leq m$, because you're reasoning on the $\pi$s and not the $\pi'$s.
In other terms you want to run out of $\pi'$s before you run out of $\pi$s with the cancellation process, and that's why you want $n \geq m$. If $n <m$ then after $n$ steps you will arrive at $1 = \pi'_{j_1} \dots \pi'_{j_{m - n}}$.
But since you can proceed the same way for the other case, that implies that $n = m$.