Is $\frac{y!}{y+1}$ an integer if $y$ is an odd number? [duplicate]

I was playing around with equations and I somehow stumbled upon the idea that if $y$ is an odd number then $\dfrac{y!}{y+1}$ is an integer. I have tried many numbers, but I do not know if I can prove it or if it is even true.

Can someone provide a proof if this is true, or a counter-example if this is false?

If $y$ is odd, then we can rewrite $y$ as $2x+1$, for some integer $x$. Then,

$$\frac{(2x+1)!}{2x+2} = \frac{1\cdot2\cdot3\cdot...\cdot x\cdot (x+1)\cdot (x+2) \cdot ...\cdot (2x+1)}{2\cdot(x+1)}$$

As you can see, you can cancel the $2$ and the $(x+1)$, leaving behind a product of integers which will, hence, be an integer.

However, for this to be true, $(x+1)>2 \Rightarrow x>1 \Rightarrow y>3$. And we get counterexamples for $y=1$ and $y=3$, as others have pointed. Hope this helps.

Edit: To clarify why the condition is required, if it does not hold, then $2$ and $(x+1)$ will not be distinct terms in the numerator but will be so in the denominator, making this line of reasoning not true.

In summary, I showed that the statement is mostly true, except for two very small counterexamples.