What is the right way to calculate a power?

You're not doing anything wrong, per se; the issue is that in order to define square root as a single-valued function on $\mathbb{R}_{\geq 0}$, we (somewhat arbitrarily) choose the positive value. So both $-1$ and $1$ square to $1$, but the square root of $1$ is only $1$, not $-1$. This becomes a bit more transparent if we replace $14$ by $2$:

$$ [(-1)^2]^{1/2} = 1^{1/2} = 1 \not= -1 = (-1)^1 = (-1)^{2/2} $$

Accordingly, the identity $(a^b)^c = a^{bc}$ does not hold generally for negative $a$, if $b$ and $c$ are not both integers.

In real numbers, the standard definition of rational exponents only permits fully reduced fractions in the exponent. Example definition from Sullivan's College Algebra:

Definition. If $a$ is a real number and $m$ and $n$ are integers containing no common factors, with $n \geq 2$, then $$a^{m/n}=\sqrt[n]{a^m}=\left(\sqrt[n]a\right)^m$$ provided that $\sqrt[n]{a}$ exists.

We have two comments about this equation:

1. The exponent $\frac mn$ must be in lowest terms and $n$ must be positive.
2. In simplifying the rational expression $a^{m/n}$, either $\sqrt[n]{a^m}$ or $\left(\sqrt[n]a\right)^m$ may be used, the choice depending on which is easier to simplify. Generally, taking the root first, as in $\left(\sqrt[n]a\right)^m$, is easier.

The first suggested transformation, $((-1)^{14})^{1/2} = 1$, is therefore invalid under this definition, because it disallows separating the numerator and denominator when they have a common factor.

Therefore only the second transformation, $(-1)^{14/2}=(-1)^7=-1$, which reduces the rational exponent first, is valid under a real-number definition like this.

The root of the problem is that $\sqrt{ 1\;}=+1$, unambiguously by definition, but $1^\frac12=(e^{2\pi i k})^{\frac12}=e^{\pi i k}\;\;\forall k\in\mathbb Z$ is not, it results to $\pm1$ depending on $k$ being odd or even.

Taking roots, you have to choose a branch, like the commonly accepted branch that $\sqrt x \ge 0$ for $x\ge0$. We could just as well have chosen the other branch, but you get in big trouble if you mix those.