Let's write the following sum in a simpler form:

$$
S_{n}=1 \cdot 1!+2 \cdot 2!+3 \cdot 3!+\ldots+n \cdot n!
$$

( $n$ ! denotes the product of all integers from 1 to $n$).
I. solution: Let's determine $S_{n}$ based on a combinatorial thought. Let's call the arrangement of $n+1$ rooks on an $(n+1) \cdot(n+1)$ chessboard, where none of them can attack each other, a free arrangement. The number of different free arrangements is clearly $(n+1)!$, because the first rook to be placed can choose the field in the first row in $n+1$ ways, and this - through its column - makes 1 field unusable in the second row, so the second rook's position can be chosen in $n$ ways, and thus the rooks in the first two rows can be arranged in $(n+1) \cdot n$ ways; the 3rd, 4th, ..., $n$-th, $(n+1)$-th rook can similarly be placed in $n-1, n-2, \ldots, 2, 1$ ways, and thus all rooks can be arranged in $F_{n+1}=(n+1) \cdot n(n-1) \ldots 2 \cdot 1=(n+1)!$ ways.

The number $F_{n+1}$ can also be determined in another way, which exhausts all free arrangements. If the first rook is placed on the 1st, 2nd, ..., $n$-th field of the first row, then outside its "attack rays" there remain $n \cdot n$ fields, on which the remaining $n$ rooks can be arranged in $n!$ ways in each case, so the number of such free arrangements is $n \cdot n!$. The same situation arises when the first rook is placed on the $(n+1)$-th (i.e., corner) field of the first row, in which case the free fields form a connected $n \cdot n$ chessboard, so the number of free arrangements on it can also be written in the form $F_{n}$, hence $F_{n+1}=n \cdot n!+F_{n}$. Similarly, for $F_{n}$, by considering the last field of the first row (the $n$-th field of the second row of the original board), we find it to be $(n-1)(n-1)!+F_{n-1}$, so $F_{n+1}=n \cdot n!+(n-1)(n-1)!+F_{n-1}$. Continuing this line of thought, we arrive at $F_{3}=2 \cdot 2!+F_{2}$, where we can already see that $F_{2}=2=1 \cdot 1!+1$, hence

$$
F_{n+1}=n \cdot n!+(n-1)(n-1)!+\ldots+2 \cdot 2!+1 \cdot 1!+1
$$

and in the first $n$ terms of this, we recognize $S_{n}$.

By counting in two ways, we get $(n+1)!=S_{n}+1$, so $S_{n}=(n+1)!-1$.

Kálmán Györy (Ózd, József Attila g. IV. o. t.)

II. solution: By a simple transformation, it directly follows that

$$
\begin{aligned}
& S_{n}=(2-1) 1!+(3-1) \cdot 2!+(4-1) \cdot 3!+\ldots+[(n+1)-1] \cdot n!= \\
& \quad=2!-1!+3!-2!+4!-3!+\ldots+(n+1)!-n!=(n+1)!-1
\end{aligned}
$$

János Bagyinszky (Nagykőrös, Arany J. g. IV. o. t.)