Sometimes I see Euler's number defined the following way:
$$e = \lim_{n \rightarrow + \infty} \biggr ( 1 + \frac 1n \biggr)^n \tag1$$
And sometimes I see the exponential function defined the following way:
$$\exp(x) = e^x = \lim_{n \rightarrow + \infty} \biggr ( 1 + \frac xn \biggr)^n \tag2$$
If we define $e$ as $(1)$, then $e^x$ becomes the following:
$$e^x = \biggr(\lim_{n \rightarrow + \infty} \biggr ( 1 + \frac 1n \biggr)^n\biggr)^x \tag 3$$
And if we define $e^x$ as $(2)$, then it remains to show that such an equation, for all $x$'s, yields a single value for $e$.
I say that because one cannot simply say $(1)$ follows from $(2)$ by definition, because I do not know whether the value for $e$ of which satisfies $(2)$ for a specific $x$ will be the same value for all $x$'s. Given $(1)$ is just the special case where $x=1$, I am not able to derive it from $(2)$ by definition, because I do not know whether the $e$-value that satisfies $(1)$ is the same $e$-value that satisfies $(2)$ for any given $x$-value.
To obtain (1) from (2), you only need to take $x=1$, as you notice.
Now, the most simply way (in my opinion) to obtain (2) from (1), is the following:
$$ \lim_{n \to +\infty} \left(1 + \frac{x}{n}\right)^n = \lim_{n \to +\infty} \left(1 + \frac{1}{\frac{n}{x}}\right)^\frac{n \cdot x}{x}$$
Fixing $x >0$ and taking $m = \frac{n}{x}$, we have that
$$ \lim_{n \to +\infty} \left(1 + \frac{x}{n}\right)^n = \lim_{m \to +\infty} \left[\left(1 + \frac{1}{m}\right)^m\right]^x = \left[\lim_{m \to +\infty}\left(1 + \frac{1}{m}\right)^m\right]^x = e^x$$
The proof for $x<0$ is similar and for $x=0$ is trivial.