As asked in the title, I am trying to arrive at the generating function
$\displaystyle\sum_{k=0}^{\infty}k!x^k$
just by starting with the generating function with constant 1 sequence $f(x) = \displaystyle\sum_{k=0}^{\infty}x^k$ and using the operation of differentiating $f(x)$ and shifting the terms in the sequence by multiplying $f(x)$ by $x$.
My attempt of the problem (which I believe is wrong).
Start with the following:
$f(x) = \displaystyle\sum_{k=0}^{\infty}x^k$
We differentiating it $k$ times and get
$f^{(k)} = \displaystyle\sum_{k=0}^{\infty}k!x^0 = \displaystyle\sum_{k=0}^{\infty}k!$
Then we shift the terms in the sequence by multiplying $x^k$ to $f^{(k)}$ and we get
$x^kf^{(k)} = \displaystyle\sum_{k=0}^{\infty}k!x^k$. This I believe is wrong because if we look at the $k^{\mathrm{th}}$ partial sum of $f(x)$, then differentiating the $k^{\mathrm{th}}$ partial term $k$ times gives us $k!$ and multiplying this by $x^k$ gives us $k!x^k$, not $\displaystyle\sum_{k=0}^{k}k!x^k$.
I understand the example of how to get from $f(x)$ to getting a generating function of sequence of squares or cubes by differentiating $f(x)$ and multiplying the result by $x$ and repeating this process once more.
If anyone can give me a perspective of how to see this problem or maybe a misunderstanding on my part, I would appreciate it.