We can formulate a recurrence relation for this as:
$$T(n) = \frac{n^2}{n-1}T(n-1)$$
Expanding using iterative method,
$$T(n) = \frac{n^2}{n-1}\times\frac{(n-1)^2}{n-2}\dots\frac{(n-k+1)^2}{n-k}T(n-k)$$.
We have the base case that $T(1) = 1$.
Therefore, $k = n-1$.
Substituting and performing basic manipulations, you should get the answer as $n!n$
So the required answer is now $\sum_{i=1}^{i=n} i \times i!$.
Note that it can be written as $\sum_{i=1}^{i=n} (i +1)i! - i!$ which is a telsecoping sum of the form $(2! - 1!) + (3! - 2!) \dots ((n+1)! - n!)$ giving the answer as $(n+1)! - 1$.