$$f_n=\Big(1+\frac{1}{nx}\Big)^n$$
$$\lim_{n \to \infty} (f_n)= e^{\frac{1}{x}}$$
so we got :
$g(x)=\Big(1+\dfrac{1}{nx}\Big)^n-e^{\frac{1}{x}}$
the derivative is therefore : $$\frac{e^{\frac{1}{x}}}{x^2}-\dfrac{\Big(1+\dfrac{1}{nx}\Big)^{n-1}}{x^2}$$
however, $e^{\frac{1}{x}}$ is always greater than $\Big(1+\dfrac{1}{nx}\Big)^{n-1}$ so derivative is always positive $\implies$ the maximum of a function is in the farthest argument $x$ (for instance in the interval $[0,1]$ it is $1$ am I correct? What happens if interval is $[1, \infty)$ , how do I determine a uniform convergence?
You are right in deducing that $g(x)$ is always increasing for $x$ positive. However, this does not mean what you think it does; notice that $g(x)$ is always negative, and so the increasing of $g$ actually implies that the difference between $f_n$ and $\exp(\frac{1}{x})$ is getting smaller as $x$ increases. The function you should be looking at is $|g(x)| = \exp(\frac{1}{x}) - f_n(x)$.
By the same reasoning you used, we see that this is always decreasing for positive $x$. Now on any interval of the form $[a,\infty)$ with $a > 0$, you will get uniform convergence: $|g(x)|$ attains its maximum at $a$, which is
$$ |g(a)| = \exp\left(\frac{1}{a}\right) - \left(1 + \frac{1}{na}\right)^n,$$
which clearly tends to $0$ as $n \to \infty$.