The question regards $p$-adic norm.
Notation:
If $|r|_{p}$ is the $p$-adic norm of $r$ then $\mu_{p}(r)$ is defined in the expression:
$|r|_{p} = p^{- \mu_{p}(r)}$.
Question:
Define: $A_{1}(m) = (2m+1) \prod_{k=1}^m (4k-1) - \prod_{k=1}^{m}(4k+1)$
Prove that, for any prime $p \ne 2$,
$\mu_{p}(A_{1}(m)) \geq \mu_{p}(m!)$.
Attempt:
From Legendre's Formula we have $\frac{m-1}{p-1} \geq \mu_{p}(m!)$.
From the strong triangle inequality: $|A_{1}(m)|_{p} \leq max(|(2m+1)\prod_{k=1}^{m}(4k-1)|_{p} , |\prod_{k=1}^{m}(4k+1)|_{p}) $
So $\mu_{p}(A_{1}(m)) \geq max(\mu_{p}((2m+1)\prod_{k=1}^{m}(4k-1)) , \mu_{p}(\prod_{k=1}^{m}(4k+1)))$
Now I attempt to find a lower bound of $\mu_{p}(A_{1}(m))$ by looking for a lower bound in the number of time $p$ can divide each product.
The number of numbers less than $4m+1$ that are divisible by $p$ at least once is $\lfloor\frac{4m+1}{p} \rfloor$. Of these numbers only $\lceil \frac{1}{2} \lfloor\frac{4m+1}{p} \rfloor \rceil$ are odd.
So we can say that $max(\mu_{p}((2m+1)\prod_{k=1}^{m}(4k-1)) , \mu_{p}(\prod_{k=1}^{m}(4k+1))) \geq\lceil \frac{1}{4} \lfloor\frac{4m+1}{p} \rfloor \rceil$.