Proposed proof for:$b! \equiv 0 \pmod a$ $\Rightarrow$ $a \le b$

Question

Let a,b $\in$ $\mathbb{Z}$. Obviously, a counterexample can disprove this statement.
I tried this approach to seek another method (possibly proof by contradiction):

$a|b!$
$a|[b(b-1)!]$

If $gcd(a,(b-1)!)=1$, then $a|b$ and hence: $a \le b$.
But I think claiming that $a|b$ is too restrictive because $a$ can be less than $b$ without having $a|b$ necessarily.

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I tried to study two cases to reach a contradiction that $gcd(a,(b-1)!)\not=1$, yet I couldn't figure out how to continue so this might actually help out :
1) $a=b$
2) $a \lt b$

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Is there any mistake that I have done? Is this proof inconclusive or are there any improvements to be made so that it is a valid proof?
Thank you in advance
EDIT: Counterexamples most certainly get the job done here. Though, I am seeking another method to disprove the statement.
EDIT 2: The original proposition is: $$a \le b \Rightarrow b! \equiv 0 \pmod a$$
I am trying to justify whether the converse of this statement is right or wrong, which is obviously wrong using counterexamples. The purpose is to find another method to disprove the converse.

Answer 1

I think there's no point to try and prove a false statement. Because $15|5!$ for example.

What is a true version of this statement is $$a|b! \implies \text{prime factors of }a \le b$$ You can try and prove this!

Update: I understood you seek to prove it without using counterexamples, so let's try:

Let $p$ and $q$ be prime numbers where $p > q$, we have $pq|p!$ while $pq >p$

Another update: in your solution of "disproving the statement" you suggested that $\gcd(a,(b-1)!)=1$ while $a \le b$. Well if $a<b$ we have $a|(b-1)!$ so this directly means $a=b$ and this is something you wouldn't want as a step in disproving the statement. So the thing you'd first try is finding $a>b$ and $a|b!$, and I've given that in my first update.

Answer 2

$b = 5$ and $a = 6$ is a counterexample.

The statement is true, however, if $a$ is further assumed to be prime.

Note that $b!$ has as factors, all natural numbers $\le b$. If $a$ is prime and $a > b$, then $a$ cannot divide $b!$ since $a$ is a prime factor of $a$ that won't divide $b!$.

If $a$ is not prime, then it is possible for all of its prime factors (with appropriate multiplicities) to be factors of $b!$ even if $b < a$, as shown above.