Let $(M, \cdot)$ be a monoid, where $|M| = k$, for some $k > 1, k \in \mathbb{N}$. The elements of this monoid have a special property: for any $a, b, x \in M$:
$$ a \cdot x = b \cdot x \Rightarrow a = b$$
Pick an element $a \in M$ such that $a \neq 1$. Define $a^0 = 1$ (is this okay to do?), and $a^n = \prod_{i = 1}^{n} a $ ($\prod_{i = 1}^{i = 2} x = x \cdot x$). Write out the following sequence:
$$ a^1, a^2, \dots, a^{k + 1}$$
By the pigeonhole principle, there must exist some $i, j \in \{1, 2, \dots, k + 1\}$ such that $a^i = a^j$ where without WLOG $i < j$. Thus we have:
\begin{align} &\quad\;\; a^i = a^j &&\text{hint: i < j, i > 0}\\ &\Leftrightarrow a^{i - 1} a = a^{j - 1} a &&\text{hint: special property of this monoid} \\ &\Rightarrow a^{i - 1} = a^{j - 1} \end{align}
If $i - 1 = 0$ (i.e. $i = 1$) then we have shown that there exists some $j \in \{1,...,k+1\}$ such that $a^j = 1$. If $i - 1 > 0$ (i.e. $i > 0$) then we can repeat the procedure:
\begin{align} &\quad\;\; a^{i - 1} = a^{j - 1} &&\text{hint: i - 1 < j - 1, i - 1 > 0}\\ &\Leftrightarrow a^{i - 2} a = a^{j - 2} a &&\text{hint: special property of this monoid} \\ &\Rightarrow a^{i - 2} = a^{j - 2} \end{align}
And we can continue to do so until we finally hit some $r \in \{1,\dots,k\}$ such that $i - r = 0$ (recall that since $i < j$, $i \in \{1,\dots,k\}$) --- since $i \leq k$, we know we can do this in less than or equal to $k$ iterations.
Thus, we have shown that any sequence generated by the powers of such a monoid must contain $1$, or put another way, for any $a \in M$, there exists $i \in {1,\cdots,k}$ such that $a^i = 1$.
This implies that $a$ has a left inverse, because:
\begin{align} &\quad\;\; a^i = 1 \\ &\Leftrightarrow a^{i - 1} \cdot a = 1 \\ &\Leftrightarrow a^{i - 1} = a^{-1} \end{align}
We can do something similar to show that (due to the associativity of $\cdot$) $a$ also has a right inverse (which also happens to be $a^{i-1}$), or we can use the fact that if $a$ has a right inverse, then its right inverse is also a left inverse (proof omitted from this discussion).
Anyway, so every $a \in M$ also has an inverse, and thus, $(M, \cdot)$ is a group.
Is this proof correct?