Estimate of variance with the lowest mean square error Regarding estimators of variance from a iid sample of size $n$,  Karl Ove Hufthammer  says Estimates of variance from an iid sample:

if they do have a normal distribution, dividing by n+1 (sic!) will give the lowest mean square error. 

Why does it give the lowest mean square error?
What does he mean by "(sic!)"? 
Thanks!
 A: Define
$$s^2_d = \frac{\sum_{i=1}^n\left(x_i-\bar{x}\,\right)^2}{d}$$
The statistic $(n - 1)s_{n-1}^2 / \sigma^2$ follows a $\chi^2_{n-1}$ distribution. A $\chi^2_{n-1}$ has mean $n-1$ and variance $2(n-1)$. Hence $\text{E}(s_{n-1}^2) = \sigma^2$, and $\text{Var}(s_{n-1}^2) = 2\frac{\sigma^4 }{ n - 1}$.
Now $s_d^2 = \frac{n-1}{d} s_{n-1}^2$
$$\text{Bias}(s_d^2) = E(s_d^2)-\sigma^2 = \frac{n-1}{d}\sigma^2 -\sigma^2=\frac{n-1-d}{d}\sigma^2$$
$$\text  {Var}(s_d^2) = 2\sigma^4(n - 1) / d^2$$
$$\text{MSE} = \text{Bias}^2 + \text{Var}$$
Hence
\begin{eqnarray}
\text{MSE}(s_d^2) &=& \left(\frac{n-1-d}{d}\right)^2\sigma^4+ 2\sigma^4(n - 1) / d^2\\
 &=&\sigma^4\frac{(n-1-d)^2+2(n-1)}{d^2}\\
&=&\sigma^4\,f(d)\,,\end{eqnarray} 
where $f(d)=1+\frac{(n-1)^2 + 2(n-1)-2(n-1)d}{d^2}$
$f(d)$ is at a turning point when $f'(d)=0$
i.e. when $d(-2(n-1))-2((n-1)^2 + 2(n-1)-2(n-1)d) = 0$
which occurs when $d-2=n-1$ ... i.e. when $d=n+1$
Showing that's a minimum rather than a maximum (or indeed a horizontal point of inflexion) is straightforward, but I'll leave it at that for now.
The relevant Wikipedia page does it more generally, getting a formula in terms of the excess kurtosis (which gives the same result in this case). I may incorporate an outline of that derivation at some point.
