I was reading this article (http://www.asiapacific-mathnews.com/03/0304/0001_0006.pdf) to see some applications of $p$-adic numbers outside mathematics, and came across these two identities:
$\sum_{n=0}^\infty (-1)^n n!(n+2) = 1$ and
$\sum_{n=0}^\infty (-1)^n n!(n^2-5) = -3$.
Since no $p$ was stated, I figured there must be some trick that shows these are the values in any $\mathbb{Q}_p$, but I couldn't find anything by just looking at some partial sums, unless I missed something. Also, where do these kinds of $p$-adic identities come from? It's easy enough to write down a series that converges in $\mathbb{Q}_p$, but writing down one with a nice limit seems a bit harder.
Before trying to prove these two you could take a look at $\sum_{n \ge 1} n\cdot n! = - 1$, an example of telescopic series. Indeed, we have $$\sum_{n =1}^m n \cdot n! = \sum_{n =1}^m (n+1)! - n! = (m+1)! - 1.$$
There is something mysterious in series of this type. For example Schikhof in his book "Ultrametric calculus" presents three more and all of them converge to an integral limit: $$ \begin{align} \sum_{n=1}^m \frac{n^2(n+1)!}{4^n} & = -4 + \frac{(2+m)(2+m)!}{4^m}, \\ \sum_{n=1}^m n^2(n+1)! & = 2 + (m-1)(m+2)! \\ \sum_{n=1}^m n^5(n+1)! & = 26 + (m^4 - m^3 - 3m^2 + 12m - 13)(m+2)!. \end{align}$$
The first one converges in $\mathbb Q_p$ exactly when $p \neq 2$, others do not require even that. What makes these odd is that you (probably) can't really replace $4$, $2$ or $5$ by any other value and preserve some nice limit. You can find the right hand sides in almost the same manner as I did with $\sum_n n \cdot n!$.
Back to your question, observe that $$ \begin{align} \sum_{n=0}^m (-1)^n \cdot n! \cdot (n^2 - 5) & = -3 + (m+1)! (m-2)(-1)^m, \\ \sum_{n=0}^m (-1)^n \cdot n! \cdot (n+2) & = 1 + (-1)^m(1+m)! \end{align}$$ But $(m+1)!$ tends to $0$ in the $p$-adic sense regardless of our choice of prime number $p$, which proves both of the identities mentioned in Rozikov's article.