[Remark: this is a gorgeous question/result]

I think you've slipped a little with your expression for $I_n$. You have $I_{0} = \frac{1}{m+1}$. But you know that $I_{n+1} = -\frac{n+1}{m+1}I_n$. So if you set $n=0$ you get $I_1 = -\frac{1}{m+1}I_0 = -\frac{1}{(m+1)^2}$. Note that I've only put $(-1)^{n-1}$ because in those $n$ terms, $I_0$ has no $(-1)$ factor, so t here are only $n-1$ factors of negative 1. Anyway, cleaning that up gives $I_n = (-1)^n \frac{n!}{(m+1)^n}$.

Can you see how this contradicts your result? Yours says that $I_1 = (-1)^1 \frac{(1+1)!}{(m+1)^{1+1}} = -\frac{2}{(m+1)^{2}}$. Which is slightly off. You haven't shown your working out for it, so I can't tell you expressed where it went wrong, but I think you've misjudged the number of terms you multiply together. Either that, or you confused finding $I_{n+1}$ with $I_n$.

We have $I_n = -\frac{n}{m+1}I_{n-1}$, agree? And $I_{n-1} = -\frac{n-1}{m+1}I_{n-2}$.

So $$I_n = -\frac{n}{m+1} \times \underbrace{-\frac{n-1}{m+1} \times \cdots I_0}_{n \, \mathrm{terms}}$$

You need to look out for the fact that you're going back to $I_0$, so the number of terms from $I_0$ to $I_{n-1}$ is $n$, not $n-1$.

Anyways, you then get $I_{n} = -\frac{n}{m+1} \times (-1)^{n-1} \frac{(n-1)!}{(m+1)^n}$. Note that I only put $(-1)^{n-1}$ because out of the $n$ terms, $I_0$ has no factor of $(-1)$ so there are only $n-1$ factors of $-1$. Cleaning this up gives $$I_n = (-1)^n \frac{n!}{(m+1)^{n+1}}$$

Which is very close to what you got, but for that crucial factor in the numerator. This should now make your final result come out alright. Let me know if any bit of this was unclear!