Chaining Exponent Rules Together I'm having trouble understanding why the following property is true and want to make sense of it before going ahead and using it in my proof by induction:
$$2^{2^n}=2^{2^{n-1}}\times 2^{2^{n-1}}=\left( 2^{2^{n-1}} \right)^2$$
I thought that the $2$ outside the bracket would multiply into $n-1$ and result in $2^{2^{2n-2}}\ldots$
Why is the result $2^{2^n}$? How does the  $-1$ in $n-1$ get cancelled out?

The subtlety you're missing in your intuition is that exponentiation is not associative; so, you clearly know the property:
$$(a^b)^2=a^{2b}$$
and you're (incorrectly) inferring that this means we could, setting $a=2^2$ and $b=n-1$, get that
$$(2^{2^{n-1}})^2=2^{2n-2}$$
however this is false because we are assuming exponentiation is associative when we ignore the implicit parenthesis; if we write this out with parenthesis, we see the error:
$$((2^2)^{n-1})^2=(2^2)^{n-1}$$
which is true, but not what we want. Impliclty, we take $a^{b^c}=a^{(b^c)}$ not $(a^b)^c$, which is what causes your error.
Rather, let $a=2$ and $b=2^{n-1}$. Then we get
$$(2^{2^{n-1}})^2=2^{2\cdot 2^{n-1}}$$
which, of course, simplifies to the correct answer
$$(2^{2^{n-1}})^2=2^{2^n}.$$
