<s>[INST] Polynomial coefficients in exponential-series: how can I convert this into a composite of $\exp(x)$? Assume we have the exponential-series $ \small \exp(x) = 1+ {x \over 1!}  + {x^2 \over 2! } + \cdots = \sum\limits_{k=0 }^\infty {x^k \over k!} $ modified with a polynomial in the coefficients
say $ \qquad \displaystyle f_1(x)= \sum_{k=0}^\infty { k^2 + k \over 2} {x^k \over k!} $
or $ \qquad \displaystyle  f_2(x)= \sum_{k=0}^\infty ( 15 k^3 + 15 k^2 - 10k - 8) {x^k \over k!} $
or in general
$ \qquad \displaystyle  f_3(x)= \sum_{k=0}^\infty ( d k^3 + c k^2 + b k + a) {x^k \over k!} $      
-: is there a good formula/algorithm/scheme how this has to be expressed as composition of the $\exp(x)$-function? (I know this can be solved using the derivatives and cancelling of k's in the polynomial with the factorials in the denominator - I'm asking for a handy/memorizable translation-formula )
Because I can factor my examples under study: is there possibly a special handy scheme, if the polynomials are given in a form like this
$ \qquad \displaystyle  f_2(x)= \sum_{k=0}^\infty ( k-1)(k-2) {x^k \over k!} $     ?
 [/INST] \begin{eqnarray}
\sum_{k=0}^\infty k\frac{x^k}{k!}&=&\sum_{k=1}^\infty \frac{x^k}{(k-1)!}=x\sum_{k=0}^\infty\frac{x^k}{k!}=xe^x;\cr
\sum_{k=0}^\infty k^2\frac{x^k}{k!}&=&x\sum_{k=1}^\infty k\frac{x^{k-1}}{(k-1)!}=x\sum_{k=0}^\infty(k+1)\frac{x^k}{k!}=x(x+1)e^x\cr
\sum_{k=0}^\infty k^3\frac{x^k}{k!}&=&\sum_{k=1}^\infty k^2\frac{x^k}{(k-1)!}=x\sum_{k=0}^\infty(k+1)^2\frac{x^k}{k!}\cr
&=&x\sum_{k=0}^\infty(k^2+2k+1)\frac{x^k}{k!}=x[x(x+1)+2x+1]e^x\cr
&=&x(x^2+3x+1)e^x.
\end{eqnarray}
Hence
\begin{eqnarray}
\sum_{k=0}^\infty (dk^3+ck^2+bk+a)\frac{x^k}{k!}&=&d\sum_{k=0}^\infty k^3\frac{x^k}{k!}+c\sum_{k=0}^\infty k^2\frac{x^k}{k!}+b\sum_{k=0}^\infty k\frac{x^k}{k!}+a\sum_{k=0}^\infty \frac{x^k}{k!}\cr
&=&dx(x^2+3x+1)e^x+c(x^2+x)e^x+bxe^x+ae^x\cr
&=&[a+bx+c(x^2+x)+d(x^3+3x^2+x)]e^x\cr
&=&[a+(b+c+d)x+(c+3d)x^2+dx^3]e^x
\end{eqnarray}
</s>