Reverse Functional Equations

My solution to this is done by induction. We first spot that $f(2) = 1$ and $\frac{f(n+1)}{f(n)} = \frac{n^2}{n-1}$ before we continue.

Base case: $n = 1$ $f(2) = 1 = 1\times f(1)$

Inductive step: Assume $f(k) = (k-1)\big( f(k-1) + f(k-2) + \dots + f(1) \big)$ for some positive integer $k \geq 2$ . Then we have $\begin{aligned} f(k+1) &= \frac{k^2}{k-1}f(k) \\ &= kf(k) + \frac{k}{k-1}f(k) \\ &= kf(k) + \frac{k}{k-1}(k-1)\big( f(k-1) + f(k-2) + \dots + f(1) \big) \\ &= kf(k) +k\big( f(k-1) + f(k-2) + \dots + f(1) \big)\\ &= k\big(f(k) + f(k-1) + \dots + f(1) \big) \end{aligned}$

Therefore by mathematical induction. We must have $f(n+1) = n\big(f(n) + f(n-1) + \dots + f(1) \big)$

It's obvious that for natural 'n' f( n + 1 ) = n(n!)

Now , Let, $[ r(r!) ] = 1(1!) + 2(2!) + 3(3!) . . . . . . . . .(n-1)(n-1!)

=    $[ r(r!) + r!  - r! ] 

=    $[ (r+1)! - r! ]

=   $[ (r+1)! ]  -  $ [ r! ]

=   [  (2!) + (3!) . . . . . (n)! ]   -  [ (1!) + (2!) + (3!) . . . . . (n-1!)]
=    (n!) - 1 ........................remember this we'll use it later.

Now, f(1) + f(2) + f(3) . . . . .f(n) = 1 + $[ r(r!) ] = 1 + [ (n)! - 1 ] = (n!)

Now, n[ f(1) + f(2) + f(3) . . . . .f(n) ] = n(n!) = f(n+1) :)