Show that $2^m \not \equiv 0 \pmod {10}$ where $m$ is a natural number.
What does this mean in relation to digits of $2^m$?
Hint: Write $m = 4q + r$ where $0 ≤ r < 4$.
We start with the Division Algorithm to write $m = 4q + r$ where $0 \le r < 4$.
And so
$$ \begin{align} 2^m & = 2^{4q+r} \\ \\ & = (2^4)^q \times 2^r \\ \\ & \equiv 6^q \times 2^r \pmod {10} \\ \\ & \equiv 6 \times 2^r \pmod {10} \end{align} $$
The last line uses the previously proven result that $6^q \equiv 6 \pmod{10}$.
We have four cases for $r$. Let's consider each in turn:
$r=0$ means $6 \times 2^r \equiv 6 \times 1 \equiv 6 \pmod{10}$.
$r=1$ means $6 \times 2^r \equiv 6 \times 2 \equiv 2 \pmod{10}$.
$r=2$ means $6 \times 2^r \equiv 6 \times 4 \equiv 4 \pmod{10}$.
$r=3$ means $6 \times 2^r \equiv 6 \times 8 \equiv 8 \pmod{10}$.
In all cases $2^m \not \equiv 0 \pmod{10}$. The means that the last digit of $2^m$ is never 0.