Prove that $p | a_i$ for some i I know there are posts regarding this  but I just wanted you guys to check my proof.
If $p$ is a prime and $p|a_1a_2...a_n$ then $p|a_i$ for some $1\leq i \leq n$.
Proof:
Note that for $n=2$, the statement holds. Assume that the statement holds for $ 1\leq n \leq k.$ For $n=k+1$, $p|a_1a_2a_3....a_ka_{k+1}$. Note that, there exists some $a_j$ such that $ 1\leq j \leq k+1$ and $gcd(p,a_j)=1$ for $j \neq i$. Then $p|a_1a_2a_3..a_{j-1}a_{j+1}..a_ka_{k+1}$. Then by induction hypothesis, $p|a_i$ for some $i \neq j$.
Edit:
Thanks to everyone who went through it. My argument was not correct.
Thanks to egreg for helping me out.
 A: Your argument is incorrect. There is no way to prove that $p$ must be coprime with some of the factors: consider the case $p=2$, $a_1=a_2=\dots=2$.
It’s much simpler: if $p\mid a_1\dots a_ka_{k+1}$, consider
$$
p\mid (a_1\dots a_k)a_{k+1}
$$
and apply what you know about the case $n=2$.
A: Its much simpler than that.
If it is true for two $a_1,a_2$ that if $p|a_1a_2$ then either $p|a_1$ or $p|a_2$ (or both).
And if it true that for any $k\ge 2$ number of $a_1, a_2,......, a_k$ that if $p|a_1a_2.....a_k$ then $p$ divides at least one of the $a_i$ (or more, possibly all, but at least one).
Then for any $k+1$ number of $a_1, a_2, ....., a_k, a_{k+1}$ then if the product $a_1a_2..... a_ka_{k+1}$ can be viewed as the product of $a_1a_2a_3 .....a_k$ times $a_{k+1}$.
That's two numbers!  so $p$ either divides $a_1a_2a_3..... a_k$ or $p$ divides $a_{k+1}$ (or both).  And if $p|a_1a_2a_3.....a_k$ it divides at least one of the $a_i; i\le k$.  So either $p$ divides at least one of the $a_i; i\le k$ or it divides $a_i; i=k+1$.  So $p$ divides at least one of to the $a_i; 1\le i \le k+1$.
So by induction the statement is true for any finite number of terms.
======= postscript ====
That wasn't the hard part.  It was supposed to be obvious.  As any product of $a_1a_2.....a_n$ of $n$ terms can be grouped into a smaller product of fewer terms it should suffice to prove it for just two terms; $a_1, a_2$.  The induction argument above is a formal prove that such a statement is valid.
However you do have to prove it is true for two terms:

Euclid's Lemmma: If $p|ab$ then $p|a$ or $p|b$ or both.

You must prove that.
=== Critique of your proof as writtenn =====
Your proof as written:

Note that for n=2, the statement holds.

Why?  That needs to be proven.

Note that, there exists some aj such that 1≤j≤k+1 and gcd(p,aj)=1 for j≠i.


*

*Why? That needs to be proven.

*It can't be proven because it isnt true.  Consider $p=7; a_1 = 35$ and $a_2=49$ and we are being asked to prove that if $7|35\times 49$ then $7|35$ of $7|49$.  You are claiming that either $\gcd(7,35) =1$ or $\gcd(7,49)=1$.  That is not true.


Then p|a1a2a3..aj−1aj+1..akak+1. Then by induction hypothesis, p|ai for some i≠j.

This isn't clear what you are claiming.  But I think you are claiming that if $p|a_1a_2...a_ka_{k+1}$ and $\gcd(a_i, p)=1$ (which you don't actually know) then $p|\frac {a_1a_2....a_ka_{k+1}}{a_i}=\underbrace{\prod\limits_{j=1;k\ne i}^{k+1}a_j}_{\text{a product of }k\text{ terms}}$
But why if $\gcd(a_i, p) =1$ would that mean $p|\frac {a_1a_2....a_ka_{k+1}}{a_i}$? That is in essence what we are being asked to prove.
That is if $p|MN$ and $\gcd(N,P)=1$ then how do you know $p|M$?  That assumes that if $p|MN$ then then $p|M$ or $p|N$ (which is what you are being asked to prove) so if $p\not \mid N$ the $p|M$.  You can't assume that until after you hve proven Euclid's lemma in the first place.
