What are the last 20 digits of mega? What are the last 20 digts of the number mega, which is
"2 in a pentagon" in steinhaus-moser-notation ?
In contrary to power towers or tetration, the ending digits
are not stable. I found out that the last digits of mega
are 56.
What about "2 in a hexagon", what are the last 20 digits of this number ?
 A: Let $T(n)$ denote number $256$ in $n$ triangles, so $T(0)=256$ and $T(n+1)=T(n)^{T(n)}$. The number mega is then equal to $T(256)$.
According to the Chinese Remainder Theorem, finding last $d\geq 1$ (decimal) digits of $T(n)$ amounts to finding $T(n)\bmod 2^d$ and $T(n)\bmod 5^d$ and a little bit of maths to combine the two.
First of all, we can make a simple observation:
$$T(n) = T(0)^{T(0)T(1)\ldots T(n-1)}$$
This immediately allows us to conclude that $2^d$ divides mega for any reasonable value of $d$. When it comes to the exponent, Euler's generalization of Fermat's Little Theorem tells us that $4^{\varphi(5^d)}\equiv 1\pmod {5^d}$ and combining $\varphi(5^d)=4\times 5^{d-1}$ and $T(0)=4^4$ gives us $$T(0)^{5^{d-1}}\equiv 1\pmod {5^d}$$ Thus, the value of the exponent is only relevant modulo $5^{d-1}$. 
These observations allow us to define a helper sequence $S(n)$ representing the product in the exponent:
$$\begin{eqnarray}
S(0) & = & 1 \\
S(n+1) & = & S(n)\cdot T(0)^{S(n)}\bmod 5^d
\end{eqnarray}$$
and get $T(n)\equiv T(0)^{S(n)}\pmod{5^d}$. A simple implementation in PARI/GP might look like this:
t(n, digits)={
    t0 = Mod(256, 5^digits);
    s = 1;
    for(k=1, n, s *= t0^(lift(s)));
    lift(chinese(Mod(0,2^digits), t0^lift(s)));
}
print(t(256, 20));

The program says last 20 digits of mega should be $49731993539660742656$. It is capable of going considerably further, though, with $1000$ being computable in a few seconds on my machine. The computation time seems to grow roughly proportionally to $d^{2.5}$.
When it comes to $2$ in hexagon, that's a considerably more dangerous beast; I'll see if and how it can be tackled to get at least some information about its digits later.
A: Since the Mega is computable by 258 exponentiations, it is easy to compute thousands of final digits of the Mega using modular arithmetic.  For larger polygons, there are too many exponentiations to compute such numbers in the same fashion.
Let us find the last 7 digits of $2$ in a hexagon.  If we define $M(m,n)$ to be $2$ in $m$ $n$-gons, then we can define $M(3,n)$ by
$M(3,0) = 2$
$M(3,n) = M(3,n-1)^{M(3,n-1)}$
Note that if $p$ is of the form $M(3,n)$, then $p$ in a square is equal to $M(3,n)$ in $M(3,n)$ triangles, which is equal to $2$ in $M(3,n) + n$ triangles.  More generally, $M(m,n)$ in an $m+1$ - gon is equal to $2$ in $M(m,n) + n \ \ \ n$-gons.  So we can define a helper function $H(m,n)$ which is defined as the number $p$ such that $2$ inside $p \ \ \ m-1$ - gons is equal to $M(m,n)$.  Then we have the recursive formula
$H(m,0) = 0$
$H(m,n) = M(m-1,H(m,n-1)) + H(m,n-1)$
and
$M(m,n) = M(m-1,H(m,n))$
Using Mathematica 10.2, I computed a table of the last $7$ digits of $M(3,n)$, and discovered that it repeated with period 3125 starting from $M(3,2)$.  So we can reduce $H(4,n)$ modulo $3125$ without affecting the last 7 digits.  This results in in $H(4,n)$ and therefore $M(4,n)$ repeating modulo $3125$, which lets us reduce $H(5,n)$ modulo $3125$ and so on.  This allows us to easily compute the last 7 digits of $M(m,n)$ for arbitrary $m,n$:
$2$ in a pentagon ends in $0742656$.
$2$ in a hexagon ends in $9341056$.
$2$ in a heptagon ends in $3395456$.
$2$ in an octagon ends in $1305856$.
$2$ in a nonagon ends in $6272256$.
$2$ in a decagon ends in $0294656$.
