What makes $5$ and $6$ so special that taking powers doesn't change the last digit? Why are $5$ and $6$ (and numbers ending with these respective last digits) the only (nonzero, non-one) numbers such that no matter what (integer) power you raise them to, the last digit stays the same? (by the way please avoid modular arithmetic) Thanks!

Modular arithmetic is the most convenient and powerful language to express the explanation in, but there is a low-tech version. The desired condition is equivalent to just $x$ and $x^2$ having the same units digit, which is equivalent to $10\mid (x^2-x)=x(x-1)$, which is equivalent to
$$10\mid x(x-1)\implies \begin{cases}10\mid x \\ 2\mid x, ~ 5\mid (x-1) \\ 5\mid x,~2\mid(x-1) \\ 10\mid (x-1)\end{cases}\implies \rm units~digit=\begin{cases}0 \\ 6 \\ 5 \\ 1\end{cases} $$

In base $b$, an integer $x$ has the same units digit no matter what power you raise it to if and only if $x$ has the same units digit as $x^2$, or equivalently $x^2\equiv x\bmod b$. This occurs when $b\mid x(x-1)$. The set of $x$ for which this occurs can be precisely characterized.
For each divisor $d\mid b$ such that $\gcd(d,b/d)=1$, the set of solutions $x$ to the system
$$\begin{cases} x\equiv 0 \bmod d \\ x\equiv 1\bmod b/d\end{cases}$$
satisfy the desired condition, and every desired $x$ is attained in this way. Using SZ aka CRT,
$$\{d\cdot\big(d^{-1}\bmod b/d\big)+bn:~d\mid b,\,\gcd(d,b/d)=1,n\in{\bf Z}\}$$
is the set of all $x$ s.t. $x,x^2,x^3,\cdots$ all have the same units digit in base $b$.
