While often accepted as a basic fact, prove $\{e,g,g^2...\}$ is a subgroup of $G$. Though often accepted as a basic fact, can it be proved that $\{e, g, g^2,\ldots ,g^k\}$ is a subgroup of $G$, where $(G, *)$ is a finite group, $g$ is an element of $G$, and $k$ is the order of $g$?  I see this as definitions everywhere, but I cannot find any proof that this is in fact, true.
To clarify, $"*"$ represents an generic binary operation, not multiplication, and $g^2 = g * g.$
 A: Let $H:= \{g^n \mid  n \in \mathbb N\}$, where $g \in G$. 
If $G$ is finite, then there exists some $k \in \mathbb N$ so that $g^k=e$, since otherwise $|G|= \infty$. So, identity is in the subgroup, and it is also enough to consider
$$H=\{g^n \mid 1 \leq n \leq k\}.$$
For each $1 \leq n \leq k$, we can take $g^n \cdot g^{k-n}=g^k=e$, but we know that $g^{k-n} \in H$, so that inverses exist.
You should show that the group is closed under multiplication.
A: First of all: if the order of $g$ is $k$, then we have that $g^k = e$, so that your group actually is $H = \{e, g, g^2, \ldots, g^{k-1}\}$. Are you aware of the subgroup criterion? (If no, it states that if $H$ is a subset of a group $G$ and $H$ is not empty, then it suffices to prove that for any $x, y \in H$ we also have $x \ast y^{-1} \in H$ in order to prove that $H$ is a subgroup. You can easily prove this yourself). 
Now clearly $H$ is not empty, since $e \in H$. Now suppose $x^l, x^m \in G$, where both $l,m \in \{0, 1, \ldots k-1\}$. Consider $x^l \ast (x^m)^{-1}$. Note that $(x^m)^{-1} = x^{-m}$ (you can prove this, but I guess this is clear). 
Now we distinguish two cases: either $l \geq m$ or $l < m$. In the first case we have that $x^l \ast x^{-m} = x^{l - m} \in G$ since $0 \leq l - m \leq k-1$. 
If $l < m$, then we would have that $x^{l-m}$ has a negative exponent. However, we have that $x^k = e$, so we find that $x^l \ast x^{-m} = x^{l-m} = x^{l-m} \ast x^k = x^{l-m + k}$ and since $l <m$, we have that $0 \leq l - m + k \leq k-1$, so that $x^{l - m + k} = x^{l-m} \in G$. 
By the subgroup criterion, we have that $H = \{e, g, g^2, \ldots, g^{k-1}\}$ is a subgroup.
