Belarus competition problems

Let us prove that $\sum_{n=1}^{2022} \frac{n^2}{(n+1)!} - \sum_{n=1}^{2022} \frac{1}{n!} < 0$, which is equivalent to the required. Note that
$$\frac{n^2}{(n+1)!} - \frac{1}{n!} = \frac{n(n+1) - 2(n+1) + 1}{(n+1)!} = \frac{1}{(n-1)!} - \frac{2}{n!} + \frac{1}{(n+1)!}$$
With this identity in mind, we make the following transformations:
$$\begin{aligned} & \sum_{n=1}^{2022} \frac{n^2}{(n+1)!} - \sum_{n=1}^{2022} \frac{1}{n!} = \sum_{n=0}^{2021} \frac{1}{n!} - 2 \sum_{n=1}^{2022} \frac{1}{n!} + \sum_{n=2}^{2023} \frac{1}{n!} = \\ & = \frac{1}{0!} - \frac{1}{1!} - \frac{1}{2022!} + \frac{1}{2023!} = \frac{1}{2023!} - \frac{1}{2022!} < 0. \end{aligned}$$