Does the following inequality hold for all $N\geq2$? \begin{equation*}\frac{\displaystyle\sum_{i=1}^{N-1}i\cdot(N-i-1)!\sum_{k=1}^{N-i}\prod_{j=k}^{k+i}j}{N(N-1)N!}<1.\end{equation*}
I have tried some numerical examples. It seems they all checked out.
If so, a proof (or any hint) will be highly appreciated.
We have $$\prod_{j=k}^{k+i}j=\frac{(i+k)!}{(k-1)!}=\binom{i+k}{i+1}(i+1)!$$
Therefore, $$ \begin{split} \sum_{k=1}^{N-i}\prod_{j=k}^{k+i}j&=(i+1)!\sum_{k=1}^{N-i}\binom{i+k}{i+1}\\ &=(i+1)!\binom{i+N-i+1}{i+2} \qquad\text{(by the hockey stick inequality)}\\ &=(i+1)!\binom{N+1}{i+2}=\frac{(N+1)!}{(N-i-1)!\cdot(i+2)}, \end{split} $$ so $$ i\cdot(N-i-1)!\sum_{k=1}^{N-i}\prod_{j=k}^{k+i}j=\frac{i}{i+2}(N+1)!=\left(1-\frac{2}{i+2}\right)(N+1)! $$ Therefore, $$ \begin{split} \sum_{i=1}^{N-1}i\cdot(N-i-1)!\sum_{k=1}^{N-i}\prod_{j=k}^{k+i}j&=(N+1)!\sum_{i=1}^{N-1}\left(1-\frac{2}{i+2}\right) \\&=(N+1)!\left(N-1-2\sum_{i=3}^{N+1}\frac{1}{i}\right)\\ &=(N+1)!\left(N-1-2\left(H_{N+1}-1-\frac{1}{2}\right)\right)\\ &=(N+1)!(N+2-2H_{N+1}), \end{split} $$ where $H_n=\sum_{i=1}^{n}\frac{1}{i}$ is the $n$th harmonic number.
Finally, $$ \begin{split} \frac{\displaystyle\sum_{i=1}^{N-1}i\cdot(N-i-1)!\sum_{k=1}^{N-i}\prod_{j=k}^{k+i}j}{N(N-1)N!}&=\frac{(N+1)!(N+2-2H_{N+1})}{N(N-1)N!}\\ &=\frac{(N+1)(N+2-2H_{N+1})}{N(N-1)}. \end{split} $$
So, your question can be restated as follows:
The equivalent bound on $H_{N+1}$ for $N\ge 2$ would be $$ 2(N+1)H_{N+1}>(N+1)(N+2)-N(N-1)=4N+2, $$ i.e. $$ H_{N+1}>\frac{2N+1}{N+1}=2-\frac{1}{N+1}, $$ or, equivalently, $$ H_n>\frac{2n-1}{n}=2-\frac{1}{n} $$ for $n\ge 3$.
And indeed, for $n\ge 3$, $$ H_n=\sum_{i=1}^{n}\frac{1}{i}=1+\sum_{i=2}^{n}\frac{1}{i}>1+\sum_{i=2}^{n}\frac{1}{n}=1+\frac{n-1}{n}=\frac{2n-1}{n}, $$ which proves the original inequality.