Here are a couple of examples, the first with almost full detail, the second with less.
First I’ll convert $156_{16}$ to base ten using repeated division in base sixteen. I’ll use $A,B,C,D,E$, and $F$ for the base sixteen digits corresponding to base ten $10,11,12,13,14$, and $15$. I’ll also use a subscript $s=16$ to indicate that a number is to be interpreted in base sixteen.
Divide $156_s$ by $A_s$. Do this just as you would in base ten: $A_s$ won’t go into $1_s$, but it will go into $15_s$. In fact $15_s=1\cdot 16+5=21$, and $A_s=10$, so it goes twice. The first digit of your quotient is $2_s$, so you need to subtract $2_s\cdot A_s$ from $15_s$.
$2_s\cdot A_s=2\cdot 10=20=1\cdot16+4=14_s$, and $15_s-14_s=1_s$, so after you bring down the $6_s$, you’re left dividing $A_s$ into $16_s$.
Similarly, $16_s=1\cdot 16+6=22$, so $A_s$ goes in twice. After you repeat the previous step (with suitable minor modifications) you have your full quotient $22_s$ and overall remainder $2_s$, as shown below.
$$\begin{array}{}
&&&2&2\\
&&\text{_}&\text{_}&\text{_}\\
A&)&1&5&6\\
&&1&4\\
&&-&-&-\\
&&&1&6\\
&&&1&4\\
&&&-&-\\
&&&&\color{red}2
\end{array}$$
Now divide $22_s$ by $A_s$. $22_s=2\cdot 16+2=34$, so the integer part of the quotient is $3_s$:
$$\begin{array}{}
&&&3\\
&&\text{_}&\text{_}\\
A&)&2&2\\
&&1&E\\
&&-&-\\
&&&\color{red}4\\
\end{array}$$
Finally, divide this last quotient, $3_s$, by $A_s$:
$$\begin{array}{}
&&0\\
&&\text{_}\\
A&)&3\\
&&0\\
&&-\\
&&\color{red}3\\
\end{array}$$
Read off the red remainders in reverse order: $156_s=342$.
Here’s one a little more complicated, the conversion of $2BA_s$ to base three.
$$\begin{array}{ccccc|cccc|cccc|cccc|ccc}
&&&E&8&&&4&D&&&1&9&&&&8&&&\color{red}2\\
&&\text{_}&\text{_}&\text{_}&&&\text{_}&\text{_}&&&\text{_}&\text{_}&&&\text{_}&\text{_}&&&\text{_}\\
3&)&2&B&A&3&)&E&8&3&)&4&D&3&)&1&9&3&)&8\\
&&2&A&&&&C&&&&3&&&&1&8&&&6\\
&&-&-&-&&&-&-&&&-&-&&&-&-&&&-\\
&&&1&A&&&2&8&&&1&D&&&&\color{red}1&&&\color{red}2\\
&&&1&8&&&2&7&&&1&B\\
&&&&-&&&-&-&&&-&-\\
&&&&\color{red}2&&&&\color{red}1&&&&\color{red}2
\end{array}$$
That last quotient of $2$ is less than the divisor, so the next division will have a $0$ quotient and remainder of $\color{red}2$, so I’ve skipped the step and colored the quotient instead. Reading the remainders in reverse order, we have $2BA_s=221212_t$ (where the subscript $t$ indicates base three).
Check: $$2BA_s=2\cdot 256+11\cdot16+10=698\;,$$ and $$221212_t=2\cdot 243+2\cdot81+1\cdot27+2\cdot9+1\cdot3+2=698\;.$$
Best Answer
Base 0 does not make any mathematical sense.
Look at binary (base 2). There are two digits, 0 and 1. Thus, every other number you need to roll over the 1 back to a zero, and add 1 to the next column.
Now, look at base 1. Now, every number requires rolling over to the next row. This is essentially a tally system, where each '1' (in base ten) gets it's own column.
Now, if you think about base 0, that would mean every increase by '1' in any non-zero base represents an infinite amount of columns that need to be created to support the overflow. Thus, every number in base 0 would essentially be infinite, or even worse, every number would be the same number.