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I am trying to convert the following recursive function to a non-recursive equation:

$$f(2) = 2$$

For $n>2$:

$$f(n)=nf(n-1)+n$$

I have calculated the results for n=2 through to n=9:

$$beginalign
f(2)&=2\
f(3)&=9\
f(4)&=40\
f(5)&=205\
f(6)&=1236\
f(7)&=8659\
f(8)&=69280\
f(9)&=623529
endalign$$

I've tried graphing the function, but have got nowhere

Any help is appreciated!

This is a linear, nonhomogenous recurrence relation. There are general methods to solve it, which you might find in the appropriate courses (I know my combinatorics text goes over it), but here's perhaps a more intuitive derivation.

Notice that you have

$$f(n) = nf(n-1) + n$$

Imagine iterating this several times: that is, we use the definition of $f$ for $f(n-1)$. We see:

$$beginalign
f(n) &= n+ nf(n-1)\
&= n + n((n-1) + (n-1)f(n-2)) \
&= n + n(n-1) + n(n-1)f(n-2)\
&= n + n(n-1) + n(n-1)((n-2) + (n-2)f(n-3)) \
&= n + n(n-1) + n(n-1)(n-2) + n(n-1)(n-2)f(n-3)\
&= ...
endalign$$

If we keep iterating this until we get to our initial condition of $f(2)=2$, then we have

$$f(n) = n + n(n-1) + n(n-1)(n-2) + ... + n(n-1)(n-2)...(3)f(2)$$

Take note: since $f(2) = 2$, the last term is actually $n!$. So what we essentially have is a sum of all of the "falling factorials" of $n$ and $n!$ itself. (A falling factorial is something like $9cdot 8 cdot 7$ - it exhibits factorial-like behavior, but doesn't go all of the way to $2$ or $1$.)

Suppose we factor out $n!$ from each term. Then we see

$$f(n) = n! left( frac1(n-1)! + frac1(n-2)! + ... + frac12! + frac11! right)$$

This begs the summation notation:

$$f(n) = n! left( sum_k=1^n-1 frac1k! right)$$

This is the summation noted in MachineLearner's answer. Then, leaning on a past MSE post, we get the sequence OP mentioned wanting a derivation of in the comments of their question - an expression involving a floor function and $e$:

$$f(n) = n! left( sum_k=1^n-1 frac1k! right) = lfloor n! cdot (e-1) rfloor - 1$$

This is derived by simply noting that $e$ has the power series

$$e = sum_k=0^infty frac1k!$$

If you start at $1$ instead, you get $e-1$ since $1/0!=1$. The summation in $f$ then becomes

$$sum_k=1^n-1 frac1k! = sum_k=1^infty frac1k! - sum_k=n^infty frac1k! = e-1 - sum_k=n^infty frac1k!$$

The remaining summation is less than $1$, and thus invites the floor function and the resulting minus one.

And thus, we conclude:

$$f(n) = lfloor n! cdot (e-1) rfloor - 1$$

the expression noted on the OEIS by Don Thousand in the comments.

The general term is given by

$$f(n)= n! sum_k=1^n-1 dfrac1k!=n!left[e-1+sum_k=n^inftydfrac1k! right]=n!(e-1)+n!sum_k=n^inftydfrac1k! .$$