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The problem is not well defined, because it is not clear what exact relation between the series and "an expression" are you looking for. The series is divergent for all $x\neq 0$. However:

It is easy to see that $y(x)=\sum_{n=0}^\infty n!x^n$ satisfies the differential equation $$x^2y'+(x-1)y+1=0,$$ and the initial condition $y(0)=1$. This is a linear first order differential equation, which can be solved explicitly: $$y(x)=x^{-1}e^{-1/x}\left(\int_x^\infty t^{-1}e^{1/t}dt+c\right).$$ Unfortunately, $c$ is arbitrary here, but your series is an asymptotic series at $0$ for all these solutions. Or "formal Taylor series", if you wish. So in some sense your series "represents" all these solutions.

For some classes of moderately diverging series the correspondence between functions and asymptotic series can be made unique. But your series is wildly divergent...

The problem is not well defined, because it is not clear what exact relation between the series and "an expression" are you looking for. The series is divergent for all $x\neq 0$. However:

It is easy to see that $y(x)=\sum_{n=0}^\infty n!x^n$ satisfies the differential equation $$x^2y'+(x-1)y+1=0,$$ and the initial condition $y(0)=1$. This is a linear first order differential equation, which can be solved explicitly: $$y(x)=x^{-1}e^{-1/x}\left(\int_x^\infty t^{-1}e^{1/t}dt+c\right).$$ Unfortunately, $c$ is arbitrary here, but your series is an asymptotic series at $0$ for all these solutions. Or "formal Taylor series", if you wish. So in some sense your series "represents" all these solutions.

For some classes of moderately diverging series the correspondence between functions and asymptotic series can be made unique. But your series is wildly divergent...

Remark. I recommend the paper by Euler, "De seriebus divergentibus", first published in Novi Commentarii academiae scientiarum Petropolitanae 5, 1760, pp. 205-237, reprinted in Opera Omnia: Series 1, Volume 14, pp. 585 - 617, Eneström-Number E247, English translation is available on Euler's web page. There he considers (among many other examples) $$\sum_0^\infty(-1)^nn!x^n$$ but this is slightly less wild than the series you proposed.

The problem is not well defined, because it is not clear what exact relation between the series and "an expression" are you looking for. The series is divergent for all $x\neq 0$. However:

It is easy to see that $y(x)=\sum_{n=0}^\infty n!x^n$ satisfies the differential equation $$x^2y'+(x-1)y+1=0,$$ and the initial condition $y(0)=1$. This is a linear first order differential equation, which can be solved explicitly: $$y(x)=x^{-1}e^{-1/x}\left(\int_x^\infty t^{-1}e^{1/t}dt+c\right).$$ Unfortunately, $c$ is arbitrary here, but your series is an asymptotic series at $0$ for all these solutions. Or "formal Taylor series", if you wish. So in some sense your series "represents" all these solutions.

The problem is not well defined, because it is not clear what exact relation between the series and "an expression" are you looking for. The series is divergent for all $x\neq 0$. However:

It is easy to see that $y(x)=\sum_{n=0}^\infty n!x^n$ satisfies the differential equation $$x^2y'+(x-1)y+1=0,$$ and the initial condition $y(0)=1$. This is a linear first order differential equation, which can be solved explicitly: $$y(x)=x^{-1}e^{-1/x}\left(\int_x^\infty t^{-1}e^{1/t}dt+c\right).$$ Unfortunately, $c$ is arbitrary here, but your series is an asymptotic series at $0$ for all these solutions. Or "formal Taylor series", if you wish. So in some sense your series "represents" all these solutions.

For some classes of moderately diverging series the correspondence between functions and asymptotic series can be made unique. But your series is wildly divergent...

The problem is not well defined, because it is not clear what exact relation between the series and "an expression" are you looking for. The series is divergent for all $x\neq 0$. However:

It is easy to see that $y(x)=\sum_{n=0}^\infty n!x^n$ satisfies the differential equation $$x^2y'+(x-1)y+1=0,$$ and the initial condition $y(0)=1$. This is a linear first order differential equation, which can be solved explicitly: $$y(x)=x^{-1}e^{-1/x}\left(\int_x^\infty t^{-1}e^{1/t}dt+c\right).$$ Unfortunately, $c$ is arbitrary here, but your series is an asymptotic series at $0$ for all these solutions.

The problem is not well defined, because it is not clear what exact relation between the series and "an expression" are you looking for. The series is divergent for all $x\neq 0$. However:

It is easy to see that $y(x)=\sum_{n=0}^\infty n!x^n$ satisfies the differential equation $$x^2y'+(x-1)y+1=0,$$ and the initial condition $y(0)=1$. This is a linear first order differential equation, which can be solved explicitly: $$y(x)=x^{-1}e^{-1/x}\left(\int_x^\infty t^{-1}e^{1/t}dt+c\right).$$ Unfortunately, $c$ is arbitrary here, but your series is an asymptotic series at $0$ for all these solutions. Or "formal Taylor series", if you wish. So in some sense your series "represents" all these solutions.