Finding a closed form of $f(n) = \frac{n-4}{n}f(n-1)+1\ ,\ n \geq 5$ where $f(4) = 1$. I'm trying to find the closed form of
$f(n) = \frac{n-4}{n}f(n-1)+1\ ,\ n \geq 5$ where $f(4) = 1$.
Empirically, it turns out the answer is simply $f(n)=1+\frac{n-4}{5}\ ,\ n \geq 4$, but I'm having a hard time getting there. I've tried two ways but neither is very successful. Could someone please suggest a way through?
Attempt 1: I try to set up a generating function
$$G(x)=f(4)x^4+f(5)x^5+\dots=\sum_{k=4}^\infty f(k)x^k$$
Substituting in the recurrence (edit: noticed I forgot to add the +1)
$$\begin{align}G(x)&=x^4+\sum_{k=5}^\infty(1-\frac{4}{k})f(k-1)x^k \\
&=x^4+x\sum_{k=5}^\infty (1-\frac{4}{k}) f(k-1)x^{k-1} \\
&= x^4+x\sum_{k=4}^\infty (1-\frac{4}{k+1})f(k)x^k  \\
&= x^4+xG(x)-x\sum_{k=4}^\infty \frac{4}{k+1}f(k)x^k
\end{align} $$
But I don't know how to express the last part in terms of G(x).
Attempt 2: Just brute force it:
With arbitrary $k$,
$$\begin{align}f(n) &= \frac{(n-4)(n-5) \dots (n-(k+3))}{n(n-1)\dots(n-(k-1))}f(n-k)\\
&\ \ + \left(1+\frac{n-4}{n}+\frac{(n-4)(n-5)}{n(n-1)}+\dots+\frac{(n-4)\dots(n-(k+2))}{n\dots(n-(k-2))}\right)\end{align}$$
Setting $n-k=4 \implies k = n -4$ and using $f(4)=1$,
$$\begin{align} f(n) &= \frac{(n-4)!}{n!/4!}+\left(1+\frac{n-4}{n}+\frac{(n-4)(n-5)}{n(n-1)}+\dots+\frac{(n-4)\dots2}{n \dots 6}\right) \\
&= {n \choose 4}^{-1}+\left(1+\frac{(n-4)!}{n!}\left(\frac{(n-1)!}{(n-5)!} + \frac{(n-2)!}{(n-6)!}+\dots +\frac{5!}{1!}\right)\right) 
 \\
&= {n \choose 4}^{-1}+\frac{(n-4)!}{n!}\sum_{k=1}^{n-5}\frac{(k+4)!}{k!} \\
&= {n \choose 4}^{-1}+\frac{1}{n(n-1)(n-2)(n-3)}\sum_{k=1}^{n-5}(k+4)(k+3)(k+2)(k+1) \\
&= {n \choose 4}^{-1}+\frac{1}{n(n-1)(n-2)(n-3)}\sum_{k=5}^{n-1}k(k-1)(k-2)(k-3)\end{align}$$
Empirically this seems to be correct, but even Mathematica refuses to simplify it all the way down. Surely there is a better way?

Hint
If you solve
$$f(n) = \frac{n-4}{n}f(n-1)+1$$ the solution is given by
$$f(n)=\frac {n+1} 5+\frac C{(n-3) (n-2) (n-1) n}$$ Then, ..., try to find the mistake.
