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 $$(2^{2^{n-1}})^2=2^{2^n}.$$