When people first devised the power notation, $x^n$ was defined only for $n = 2,3,4,\dots$ meaning $n\; x's$ multiplied together. From that definition certain rules were obtained, such as $a^n \times b^n = (ab)^n,$ and
$a^m \times a^n = a^{m+n}.$

Extending this to defining $x^1 = x$ was straightforward, even though there is no multiplication involved. Using this definition, all the rules still work. Clearly $a^1 \times b^1 = ab = (ab)^1;$ and more importantly

$\begin{eqnarray}a^1 \times a^n & = & a \times (a \times a \times \dots \times a, \mbox{ n times })\\
& = & a \times a \times \dots \times a, \mbox{ n+1 times }\\
& = & a^{n+1}.\end{eqnarray}$

Now we get curious and ask about $x^0.$ We need $a^0 \times a^n = a^{0 \; + \; n} = a^n.$ This forces
$a^0 = 1$ (answering your question), unless $a=0$ in which case $a^n = 0$ and any value of $a^0$ would work. In fact, it turns out that no value of $0^0$ works really well, and it is better to leave it undefined like $\large \frac{0}{0}.$

Using the same idea - extend the definition while keeping the same properties - we can define negative powers (which are usually fractions), fractional powers (which are usually irrational and sometimes complex) and irrational and complex powers. We end up with strange and wonderful results like $e^{\pi i} = -1$ (where e is the base of the natural logarithms and i is the square root of -1)!

Good Hunting!
RD