Alternate form of $\operatorname {Ei}(x) = \int_{-\infty}^{x}\frac{e^t}{t}dt$

Question

By definition, Ei$(x)$ is defined by $$\text{Ei}(x)=\int_{-\infty}^{x}\frac{e^t}{t}dt,\text{ where }x\in{\mathbb{R}}{\setminus}{\{}0{\}}$$ Integrating by parts multiple times, using $\frac{dv}{dt}=e^t$ and u being the rest of the integrand (for example, $\frac{1}{t}$, then $\frac{-1}{t^2}$, etc), gives the following;

$$\text{Ei}(x)= \bigg[\frac{e^t}{t}\bigg]_{\infty}^{x}+\int_{-\infty}^{x}\frac{e^t}{t^2}dt$$ $$=\bigg[\frac{e^t}{t}+\frac{e^t}{t^2}\bigg]_{\infty}^{x}+2\int_{-\infty}^{x}\frac{e^t}{t^3}dt$$ $$=\bigg[\frac{e^t}{t}+\frac{e^t}{t^2}+2\frac{e^t}{t^3}\bigg]_{-\infty}^{x}+(3*2)\int_{-\infty}^{x}\frac{e^t}{t^4}dt$$ $$=\bigg[\frac{e^t}{t}+\sum_{k=1}^{n}k!\frac{e^t}{t^{k+1}}\bigg]_{-\infty}^{x}+n!\int_{-\infty}^{x}\frac{e^t}{t^{n+1}}dt$$ my question is;

When we consider the case of $n\to\infty$, what happens to the final integral? My intuition says that $$\lim_{n\to\infty}\bigg(n!\int_{-\infty}^{x}\frac{e^t}{t^{n+1}}dt\bigg)$$ can simply not be evaluated? If the integral goes to zero, then we would have $$\text{Ei}(x) = \bigg[\frac{e^t}{t}+\sum_{k=1}^{n}k!\frac{e^t}{t^{k+1}}\bigg]_{-\infty}^{x}= \frac{e^x}{x}+\sum_{k=1}^{n}k!\frac{e^x}{x^{k+1}}-\lim_{R\to-\infty}\bigg(\frac{e^R}{R}+\sum_{k=1}^{n}k!\frac{e^R}{R^{k+1}}\bigg)=$$ $$\frac{e^x}{x}+\sum_{k=1}^{n}k!\frac{e^x}{x^{k+1}}$$ which is a form of the exponential integral function which i have never come across before, making me believe that it is incorrect. However, if the aforementioned integral does not tend to zero as $n\to\infty$, then i am at a loss for how to proceed.