A chi-square test is the first thing that comes to mind:
$$
\sum\frac{(\text{observed} - \text{expected})^2}{\text{expected}}
$$
If you roll the die $n$ times, the "expected" number of times you would see any particular outcome is $n/6$. If $n$ is large, this has approximately a chi-square distribution with 5 degrees of freedom. You reject the null hypothesis of fairness if the test statistic given above is large.
95% confidence does mean one out of twenty fair dice will fail.
See also this amazing analysis by a physicist of perhaps the most extensive experiment of this kind ever done: http://bayes.wustl.edu/etj/articles/entropy.concentration.pdf
A further refinement of the chi square test would be to note that each outcome of a roll has an opposite face. If one outcome is unusually high, the opposite face should be unusually low. Thus, it is the difference between opposite face frequencies that detect unbalance in the die. You could create a simulation in a simple spreadsheet and find the confidence limits by Monte Carlo.
Starting with your last questions first: yes, the $\chi^2$ distribution with $k$ degrees of freedom is normally defiined as being the sum of the squares of $k$ independent $N(0,1)$ distributions.
For the moment generating function, note that since the MGF of a sum of independent variables, is the product of the MGFs, if $M_k$ denotes the moment generating function of $\chi^2(k)$ then
$$M_k(s) = M_1(s)^k.$$
Then to derive $M_1(s)$, denoting $X = N(0,1)^2$ and $f$ for the pdf of a $N(0,1)$ variable
\begin{align*}
M_1(s) & = \mathbf E \left[ e^{sX^2} \right] \\
& = \int_{-\infty}^\infty \exp(sx^2) f(x) dx \\
& = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \exp\left(sx^2\right) \exp\left(-x^2/2\right) dx\\
& = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \exp\left((s-1/2)x^2\right) dx\\
& = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \exp\left(-\frac{x^2}{2 (1-2s)^{-1}}\right) dx\\
& = \frac{\sqrt{2 \pi (1-2s)^{-1}}}{\sqrt{2 \pi} }\\
& = \left(1 - 2s\right)^{-\frac12}
\end{align*}
where in the second to last line we used the fact that the integral is the un-normalized pdf of a $N\left(0, (1-2s)^{-1} \right)$ distribution.
From the above we then get
$$ M_k(s) = M_1(s)^k = (1 - 2s)^{-k/2}.$$
Why is the MGF of a sum of independent variables the product of their MGFs?
This follows from the fact that given independent random variables $X,Y$ and functions $U,V$ then
$$ \mathbf E [ U(X) V(Y)] = \mathbf E[U(X)] \mathbf E[V(Y)]$$
In the special case that $U_s(x) = V_s(x) = \exp(sx)$ then
$$U_s(X)V_s(Y) = \exp(sX)\exp(sY) = \exp(s(X+Y)),$$
so the result about the MGFs follows.
To justify the general claim, we note that the joint distribution function of two independent variables satisfies $f_{X,Y}(x,y) = f_X(x)f_Y(y)$ and then
\begin{align*}
E [ U(X) V(Y)] &= \int_{-\infty}^\infty \int_{-\infty}^\infty U(x) V(y) f_{X,Y}(x,y) dx dy \\
& = \int_{-\infty}^\infty \int_{-\infty}^\infty U(x) f_X(x) V(y)f_Y(y) dx dy \\
& = \int_{-\infty}^\infty U(x)f_X(x) \left(\int_{-\infty}^\infty V(y)f_Y(y)dy \right) dx \\
& = \int_{-\infty}^\infty U(x)f_X(x) \mathbf E[V(Y)] dx\\
& = \mathbf{E}[V(Y)] \int_{-\infty}^\infty U(x)f_X(x) dx\\
& = \mathbf E[V(Y)] \mathbf E[V(X)]\\
\end{align*}
Best Answer
The reason you only have $k-1$ degrees of freedom is that if you roll a $k$-sided die $100$ times, you get $k$ different counts (the number of times you roll $1$, the number of times you roll $2$, and so on) but not all $k$ of them can be chosen separately. Once you have chosen $k-1$, I can tell you the $k^{\text{th}}$: it's just whatever is left over from $100$.
(Equivalently, if we divide through by $100$, or whatever the total number of samples was, you get $k$ proportions: but one of them can be deduced from the others, because all $k$ proportions together must add up to $1$.)
In general, the number of degrees of freedom in a $\chi^2$ test is the number of proportions that need to be specified before the others can be deduced.
So yes: if the die has $6$ sides, then the number of degrees of freedom is $5$. If you tell me that $10\%$ of your rolls were $1$s, $10\%$ were $2$s, $10\%$ were $3$s, $20\%$ were $4$s, and $20\%$ were $5$s, I can deduce that $30\%$ were $6$s: there is no freedom in specifying the last percentage.