Prove a sequence converges to 1

I am new to the world of proofs and I'm working on an assignment for Analysis. While I understand the concept of convergence I'm finding that writing the proofs is difficult for me and I want to ask if my logic is sound here and what I can improve on.

Problem: For each positive integer $\displaystyle n$, let $\displaystyle p_{n} = 1 - \frac{1}{n}$. Show that the sequence$\displaystyle p_{1}, p_{2}, p_{3}, ...$ converges to 1.

Proof:

To show that $\displaystyle p$ converges to 1, we must show that if $\displaystyle S$ is an open interval containing 1, then $\displaystyle \exists$ a positive integer $\displaystyle N$ such that if $\displaystyle n$ is a positive integer and $\displaystyle n >= N$, then $\displaystyle p_{n} \in S$. Let $\displaystyle S$ be an open interval containing 1 and let $\displaystyle n \in \mathbb{Z}^+ $and $\displaystyle N \in \mathbb{Z}^+$. Let $\displaystyle S$ be the interval $\displaystyle (1-\epsilon, 1+\epsilon)$, $\displaystyle \epsilon > 0$. Then $\displaystyle |p_{n} - 1| < \epsilon$ in order for $\displaystyle p_{n} \in S$. Since $\displaystyle p_{n} = 1 - \frac{1}{n}$, $\displaystyle |p_{n} - 1| = |1 - \frac{1}{n} - 1| = |-\frac{1}{n}| = |-1 * \frac{1}{n}| = |\frac{1}{n}|$. And $\displaystyle |\frac{1}{n}| < \epsilon$. Since $\displaystyle n$ is positive, we can say $\displaystyle \frac{1}{n} < \epsilon \Longrightarrow \frac{1}{\epsilon} < n$. Now let $\displaystyle N = \frac{1}{\epsilon}$. Since for all $\displaystyle n > N p_{n} \in S$, $\displaystyle p$ converges to 1.

A few things I'm concerned about are at the end when I let $\displaystyle N = \frac{1}{\epsilon}$; if I say this is it implied that $\displaystyle \frac{1}{\epsilon}$ is a positive integer, or can that not be assumed because we don't know that 1 evenly divides $\displaystyle \epsilon$? Also, I showed for $\displaystyle n > N$ instead of $\displaystyle n >= N$ but because of the way I formed the interval I'm not sure how to fix that. Any tips would be greatly appreciated!