I have found a moment generating function $M_n$ given by
$\cfrac{(1-e^t)e^{\frac tn}}{n(1-e^{\frac tn})}$ if $t\ne 0$ and 1 if $t =0$
How do I prove that $M_n$ converges point-wise to the moment generating function of a random variable that is uniformly distributed on $(0,1)$?
The moment generating function of a uniform random variable on $[a,b]$ is given by \begin{align*} M(t)=\begin{cases}\frac{e^{tb}-e^{ta}}{t(b-a)}&\text{if }t\neq0;\text{ and}\\ 1&\text{if }t=0.\end{cases} \end{align*} Thus, your problem amounts to showing that $$\lim_{n\to\infty}\frac{(1-e^t)e^{t/n}}{n(1-e^{t/n})}=\frac{e^t-1}{t}$$ whenever $t$ is a fixed real number other than $0$. We can break this problem down in several sub-problems:
First, we find $\lim_{n\to\infty}(1-e^t)e^{t/n}$ for some $t\neq 0$. Let $t\neq 0$ be fixed. Clearly, $t/n$ converges to zero as $n\to\infty$. Thus, since the function $x\mapsto e^x$ is continuous, it follows that $e^{(t/n)}\to e^0=1$ as $n\to\infty$. Since $(1-e^t)$ is constant, we conclude: $$\lim_{n\to\infty}(1-e^t)e^{t/n}=1-e^t.$$
Second, we find $\lim_{n\to\infty}n(1-e^{t/n})$. If we try this directly, we obtain $\infty\cdot 0$, which is an indeterminate form. To solve this, we can use a clever trick: $$\lim_{n\to\infty}n(1-e^{t/n})=\lim_{n\to\infty}\frac{1-e^{t/n}}{1/n},$$ which gives $0/0$. This is still an indeterminate form, but you can apply l'Hôpital's rule to get rid of the indetermination. This should give you that $$\lim_{n\to\infty}\frac{1-e^{t/n}}{1/n}=\lim_{n\to\infty}-te^{t/n}=-t.$$
Finally, use the fact that, if two limits $\lim_n x_n$ and $\lim_n y_n$ exist and are finite, then $$\lim_{n\to\infty}\frac{x_n}{y_n}=\frac{\lim_{n\to\infty}x_n}{\lim_{n\to\infty}y_n}.$$