How to prove the product of first n consecutive odd numbers is a square less than another square? I have observed for first few values of consecutive odd numbers, the result is always of the form:
$m^2 - n^2$, where $m$ and $n$ are two distinct positive integers. That is:
$$1\cdot 3\cdot 5 \cdot 7\cdots (2k-1) = m^2 - n^2$$
For example:
$
1\cdot 3 = 3 = 2^2 - 1^2 \\
1\cdot 3\cdot 5 = 15 = 4^2 - 1^2 \\
1\cdot 3\cdot 5\cdot 7 = 105 = 11^2 - 4^2 \\
1\cdot 3\cdot 5\cdot 7\cdot 9 = 945 = 31^2 - 4^2 \\
\vdots
$
But not sure, how to prove it.
Here is an attempt using induction:
Let it be true for some value of k, that is:
$1\cdot 3\cdot 5\cdot 7\cdot 9\cdots(2k - 1) = m^2 - n^2$
Then when $k$ takes the value of $k+1$, we have
$$\begin{align}
1\cdot 3\cdot 5\cdot 7\cdot 9\cdots(2k - 1)\cdot(2k + 1)
&= (m^2 - n^2)\cdot(2k + 1)\\
&= (m^2 - n^2)\cdot {(k+1)^2 - k^2}
\end{align}$$
and got stuck here.
Can you please suggest to proceed further or an altogether a different way of proving so or prove me wrong.
Thanks in advance.

You might note that the difference sequence of the sequence of squares is exactly the odd numbers.  Here's the sequence of squares:
$$0, 1, 4, 9, 16, 25, 36, \ldots$$
Here's the sequence of the differences:
$$1, 3, 5, 7, 9, 11, \ldots$$
That means every odd number is the difference of two consecutive squares.  And your number is odd.
