The problem given is as follows: (from CSAT, here, pg 3, Q11)
An organism is born on day $k = 1$ with 1 cell. During day $k = 2, 3,\dots$ the organism produces $\frac{k^2}{k−1}$ times more new cells than it produced on day $k − 1$. Give a simplified expression for the total of all its cells after $n$ days. [HINT:This is different to the number of new cells produced during day $n.$]
My attempt:
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, I got the answer as $n!n$.
So what we are required to calculate is $\sum_{i=1}^{n}i! \times i$.
However, this is where I got stuck. Is this the most simplified version or can it be further solved?
Hint: You've done almost all of the work and as I see you are preparing for an exam:
The first few terms (for $n=1,2,3,4$) are $$1,5,23, 119$$
Do those numbers look familiar if you add $1$?
Induction should do the job.