How do I expand this exponential limit using Taylor expansion?

I have this equation:

$$\lim_{n\to \infty}\left(1+{x\over n} \right)^n=e^x$$

I need to Taylor expand both sides to prove that they are equal. Now I know how to do this using l'Hopital's rule, but I'm not supposed to prove this in that way. This is what I've tried so far:

$$\lim_{n\to \infty}\left(1+{x\over n} \right)^n$$
$$=\lim_{n\to \infty}e^{n\big[\log\big( 1+{x\over n}\big)\big]}$$
$$=\lim_{n\to \infty}\big(1+n\log\big(1+{x\over n}\big)+{n^2\over 2!}{\log^2\big(1+{x\over n}}\big)+O(x^3)\big)$$

This doesn't look anything like the expansion for $e^x$, so I tried to pass the limit through the exponential, expand $\log_{n\to \infty}$, and solve it that way:

$$e^{\lim_{n\to \infty}\big(1+{x\over n} \big)^n}$$
$$e^{\lim_{n\to \infty}\big[1+n\big({x\over n}-{x^2\over 2n^2}+{x^3\over 3n^3}-O(x^4)\big) \big]}$$

Distributing the $n$ into the parentheses, and taking the limit of my result was this:

$$e^{\lim_{n\to \infty}\big(1+{x\over n} \big)^n} = e^{1+x}=e^1e^x$$

And it looks very close to what I need it to be. I also have the problem that there aren't any factorials in the denominators of my expansion. I found this question: Limit Question involving logarithmic taylor expansion. However, it didn't help me very much.

How should I expand this equation and what am I doing wrong?

Answer

In fact with taylor expansion of $\ln(1+x)$ at $x=0$ you have
$$
\ln(1+x)=x-\frac{x^2}{2}+o\left(x^2\right)

$$
So with $\frac{x}{n}\underset{n \rightarrow +\infty}{\rightarrow}0$ you can use it and

$$e^{n\ln\left(1+\frac{x}{n}\right)}=e^{n\left(\frac{x}{n}+o\left(\frac{1}{n}\right)\right)}=e^{x+o\left(1\right)}\underset{n \rightarrow +\infty}{\rightarrow}e^{x}$$