Eleven times eleven times eleven

It is given that $N = p^p$ has $1331$ digits. This means that $1330 \le \log{p^p} < 1331$ $\Rightarrow 1330 \le p\log{p} < 1331$ .

By trial and error, we note that $500 \log {(500)} = 1349.48$ and $490 \log {(490)} = 1318.20$ . Therefore, $490 < p < 500$ . Calculating the values of $x \log {x}$ for $490 < p < 500$ , we have:

$\begin{aligned} 491 \log {(491)}= 1321.321013 \\ 492 \log {(492)}= 1324.446831 \\ 493 \log {(493)}= 1327.573531 \\ 494 \log {(494)}= 1330.701113 \\ 495 \log {(495)} = 1333.829573 \end{aligned}$

So we note that only $1330 \le 494 \log{(494)} < 1331$ , therefore $p = \boxed {494}$ .

p^p has 1331 digits.
log (p^p)=1330+k (0<k<1)
So, p=494.

Python 2.7:

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n = 1
while len(str(n**n)) != 1331:
    n += 1
print "Answer:", n