Here's how I think of your result:

Let's look for integers $n$, such that the beginning of the decimal expansion of $n^{\log n}$ agrees with that of $\pi$ (up to some point). Using a for loop, I found the following approximations for $n<100,000$:

$$ \pi \approx \frac{11^{\log 11}}{10^2},\frac{53599^{\log{53599}}}{10^{51}},\frac{59546^{\log{59546}}}{10^{52}}.$$

Note that the last two only approximate $\pi$ to 4 digits after the decimal point.

It seems that for $n<1,000,000$, $n=11$ gives the best approximation of the form $\frac{n^{\log{n}}}{10^{d(n^{\log n})-1}}$ where $d(m)$ is the number of digits of $m$ left to the decimal point.

I'm stil trying to find better approximations though...

**EDIT:**

In Mathematica I used something of the form

```
For[n = 1, n < 100000, n++,If[Floor[n^Log[n]/10^(IntegerLength[Floor[n^Log[n]]] - 5)] == 31415,Print[N[n^Log[n], 10], " ", n]]]
```

This will give you approximations good to 4 decimal places in the range $n<100000$.

**EDIT2:**

Using a longer loop for finer approximations I found

$$\pi \approx \frac{3214471^{\log 3214471}}{10^{97}},\frac{3745521^{\log 3745521}}{10^{99}} $$

both to 6 decimal places.

Overall, it seems that the case $n=11$ is extraordinarily good for small values of $n$. I still can't see why nonetheless.