Prove that $\lim_{n\to \infty} (1+\frac{1}{n^3})^n = 1$ using definition of limit I cannot seem to understand how to prove $\lim_{n\to \infty} (1+\frac{1}{n^3})^n = 1$ through the usual $|a_n-l|<\epsilon$. I must do this using no other theorem, just the definition. 
Doing it for $\lim_{n\to \infty} (1+\frac{1}{n^3}) = 1$ is really easy, but I do not know how to handle the exponentiation.
Thank you very much for any help.
EDIT: An example of proof would go like this:
$\epsilon>0, \forall N>n$
$|1-\frac{1}{n^3}-1|=|-\frac{1}{n^3}|=\frac{1}{n^3}<\frac{1}{n}<\frac{1}{N}<\epsilon$
$N>\frac{1}{\epsilon}$
Clarification:
$\epsilon>0, \forall N>n$
I need to find what is the value of $N$ in relation to $\epsilon$ when:
$|(1-\frac{1}{n^3})^n-1|<\epsilon$ (the only manipulation allowed being the ones I showcased in my example).
 A: By Bernoulli's inequality for $n\geq 2$ $$\left(1-\frac1{n^3+1}\right)^n\geq 1-\frac{n}{n^3+1}>1-\frac1{n^2}$$ and taking reciprocals $$\left(1+\frac1{n^3}\right)^n < \frac1{1-\frac1{n^2}} = 1+\frac1{n^2-1}$$
A: Elaborating on lisyarus comment:
$$
\left(1+\frac{1}{n^3}\right)^n=\sum_{k=0}^n {{n}\choose{k}}\frac{1}{n^{3k}}
$$
and therefore
$$
\left|\left(1+\frac{1}{n^3}\right)^n-1\right|=\sum_{k=1}^n {{n}\choose{k}}\frac{1}{n^{3k}}=R(n)
$$
Now, you need to find $N$ such that $n>N$ implies $R(n)<\varepsilon$. This can be done using the fact that
$$
{{n}\choose{k}}=\frac{n(n-1)\cdots(n-k+1)}{k!}\leq \frac{n^k}{k!}.
$$
But then
$$
R(n)\leq \sum_{k=1}^{n}\frac{n^k}{k!}\frac{1}{n^{3k}}=\sum_{k=1}^{n}\frac{1}{k!n^{2k}}\leq \sum_{k=1}^{n}\frac{1}{n^2}=\frac{1}{n}.
$$
Therefore, if one takes $N=1/\varepsilon$, we have that $n>N$ implies $R(n)\leq\varepsilon$ (strictly if $\varepsilon<1$).
A: Let $$a_n=(1+\frac{1}{n^3})^n$$
Given an $\epsilon>0$,
We have
$$ln(a_n)=n ln(1+\frac{1}{n^3})$$
$$=n(\frac{1}{n^3}(1+\epsilon(n))),$$
with $\lim_{n\to\infty}\epsilon(n)=0$.
this gives
$$ln(a_n)=\frac{1}{n^2}(1+\epsilon(n))$$
and
$$a_n=e^{\frac{1}{n^2}(1+\epsilon(n)) }$$
or
$$a_n=1+\frac{1}{n^2}(1+\epsilon(n))$$
which implies
$$|a_n-1|=\frac{1}{n^2}(1+\epsilon(n)).$$
$\exists N_1 \in \mathbb N $ such that 
$$n\geq N_1 \implies -1<\epsilon(n)<1$$
and $N_2$ such that $n\geq N_2 \implies  \frac{2}{n^2}<\epsilon$ given by
$N_2=\lfloor \sqrt{\frac{2}{\epsilon}}\rfloor +1$.
at the end, we take $N=max(N_1,N_2)$
we check that
$$n\geq N \implies |a_n-1|<2\frac{\epsilon}{2}.$$
