limit in p-adic number system Please give me a hint for limit of $lim_{n\to\infty}3^{2^n}$ in $\mathbb Q_5$. 
First, new absolute value $|\cdot|'$ on $\mathbb Q$ is defined as the following:
For $\frac{n}{m}\in \mathbb Q$,$$|\frac{n}{m}|'=5^{-v}$$ 
if $\frac{n}{m}=5^vp_1^{v_1}p_2^{v_2}p_3^{v_3}...p_r^{v_r}$ where $p_i$: distinct prime number without 5 and $v_i\in\mathbb Z$.
Then, the absolute value satisfies $|r|'=0 
 \Leftrightarrow r=0$ and triangle inequiality. Secondly, based on this absolute value and equivalence relation for Cauchy sequence, new $\mathbb Q_5$
system is constructed. 
 A: By repeated squaring we have $3^4\equiv 6\bmod 25$, $3^8\equiv 11$, $3^{16}\equiv 21$, $3^{32}\equiv 16$, $3^{64}\equiv 6$ so the expression cycles $\bmod 25$.  The terms in the cycle differ by amounts having $5$-adic norm $1/5$ which never goes to zero.  So there cannot be a $5$-adic limit.
A: This is a moderately delicate question. Let’s look at $3^{2^{n+2}}$ modulo $25$. We have:
$$
3^{2^{n+2}}=3^{4\cdot2^n}=81^{2^n}=(1+5\cdot16)^{2^n}\equiv(1+5)^{2^n}\equiv1+2^n\cdot5\pmod{25}
$$
Now, you can check that for $n\equiv0,1,2,3\pmod4$, you get $2^n\equiv1,2,4,3\pmod5$ respectively, in other words, $3^{2^{n+2}}\equiv6,11,21,16\pmod{25}$ respectively, so there’s nothing like a limit here.
In fact the situation is even worse (or better, depending on your taste). Using the $5$-adic logarithm, you can show that when you restrict to values of $n\equiv0\pmod4$, the numbers $(1+5\cdot16)^{2^n}$ are dense in $6+25\Bbb Z_5$, and similarly for $n\equiv1\pmod4$, the numbers $(1+5\cdot16)^{2^n}$ are dense in $11+25\Bbb Z_5$, etc. But this density question involves more work than you want to see now.
The moral is: no limit.
A: Since you're repeatedly squaring in the limit, if the limit exists,
$$\lim_{n\to \infty} 3^{2^n} = L$$
then it must end up satisfying $L^2=L$. So either L=0 or L=1. It's not 0 because 3 is not divisible by 5, so it must be 1 if it exists. That means this limit must be zero:
$$\lim_{n \to \infty} 3^{2^n} - 1 = 0$$
We can factor that pretty easily by a geometric series, for reference:
$$\frac{3^{2^n}-1}{3-1} = \sum_{k=0}^{2^n-1} 3^k$$
so we can rewrite the previous limit as,
$$\lim_{n \to \infty} 2*\sum_{k=0}^{2^n-1} 3^k = 0$$
in the limit if this sum converges (it doesn't), it should converge to:
$$2 * \frac{1}{1-3}$$
which is not 0, so the limit does not exist.
