How prove this $3^{\frac{5^{2^n}-1}{2^{n+2}}}\equiv (-5)^{\frac{3^{2^n}-1}{2^{n+2}}}\pmod {2^{n+4}}$ Question:

show that:
$$3^{\frac{5^{2^n}-1}{2^{n+2}}}\equiv (-5)^{\frac{3^{2^n}-1}{2^{n+2}}}\pmod {2^{n+4}},n\geq 1$$

My idea: since
I have prove
$$5^{2^n}-1\equiv 0\pmod {2^{n+2}}$$
$$3^{2^n}-1\equiv 0\pmod {2^{n+2}}$$
so let
$$\frac{5^{2^n}-1}{2^{n+2}}=k_{1},\frac{3^{2^n}-1}{2^{n+2}}=k_{2},k_{1},k_{2}\in N$$
so we must prove
$$3^{k_{1}}-(-5)^{k_{2}}\equiv 0\pmod {2^{n+4}}$$
and How prove it?
Thank you
 A: We can check with an easy induction that the orders of $3$ and $-5$ modulo $2^{n+4}$ are $2^{n+2}$, so in order to check the equality it is enough to know $\frac {a^{2^n}-1}{2^{n+2}}$ modulo $2^{n+2}$, hence $a^{2^n}-1$ modulo $2^{2n+4}$ (where $a=3$ and $5$)
Suppose $a$ is odd and define for $n\ge 1$ the sequence $(u_n)$ by $a^{2^n} = 1 + u_n.2^{n+2} \pmod {2^{2n+4}}$. Call $v_n.2^{2n+4}$ the error term in the modulo.
Then $a^{2^{n+1}} = (a^{2^n})^2 = 1 + u_n.2^{n+3} + v_n.2^{2n+5} + u_n^2.2^{2n+4} \pmod {2^{2n+6}}$, so $u_{n+1} = u_n + (v_n \pmod 2).2^{n+2} + (u_n^2 \pmod 4).2^{n+1}$
This tells you that when $u_1$ is odd (this is equivalent to $a \equiv 3,5 \pmod 8$), to get $u_{n+1}$ you have to flip the last bit of $u_n$ and add a "random" bit of weight $2^{n+2}$.
So the sequence $u_n$ converges to a $2$-adic integer $u(a)$, and $u_n$ is $u(a) \pmod {2^{n+2}}$ with its last bit flipped.
You can even drop the modulo and deduce that the sequence $\frac 14 (a^{2^n}-1)/2^n$ converges $2$-adically to $u(a)$

Remember that for any integers $x$ and $y$, $(1+x)^y = 1 + yx + \frac{y(y-1)}2 x^2 + \frac{y(y-1)(y-2)}6 x^3 + \ldots $
Using an even $x$ and letting $y=2^n$,
the sequence $((1+x)^y - 1)/y = x + \frac{y-1}2x^2+\frac{(y-1)(y-2)}6x^3 + \ldots$
 converges $2$-adically to $x - x^2/2 + x^3/3 - x^4/4 + x^5/5 - \ldots$
which we will call $\log(1+x)$
Going back to our $u$ and putting $x = a-1$ (which is even), we deduce that $u(a) = \frac 1 4 \log (a)$.

So far we have determined that for $a \equiv 3,5 \pmod 8$, $\frac{a^{2^n}-1}{2^{n+2}} \pmod {2^{n+2}} = \frac 14 \log(a) + 2^{n+1} \pmod {2^{n+2}}$.
Since $3$ and $-5$ are of order $2^{n+2}$ modulo $2^{n+4}$, their $2^{n+1}$th power has to be $-1$, so we can simplify your question to :
Is $3^{\frac 14 \log 5} \equiv (-5)^{\frac 14 \log 3} \pmod {2^{n+4}}$ ?
Actually you can define $x^y$ by the formula above in the $2$-adics as long as $x$ is odd, and you can define $\exp(x) = \sum \frac {x^n}{n!}$ when $x$ is a multiple of $4$.
They satisfy the useful relations $\exp(\log(x)) = x$ when $x \equiv 1 \pmod 4$, $(x^z)(y^z) = (xy)^z$, $\exp(x)^y = \exp(xy)$, $\log(xy) = \log(x)+\log(y)$.  
In particular, $\log 1 = 0$, then $\log (-1) = 0$, and so $\log x = \log (-x)$, and $\exp(\log x) = -x$ when $x \equiv 3 \pmod 4$.
Finally, $(-1)^x = (-1)^{x \pmod 2}$

Then $3^{\frac 14 \log 5} = (-1)^{\frac 14 \log 5} \exp(\frac 14 \log 5 \log (-3)) = - \exp(\frac 1 4 \log 5 \log 3)$
And $(-5)^{\frac 14 \log 3} = (-1)^{\frac 14 \log 3} \exp (\frac 14 \log 5 \log 3) = - \exp(\frac 1 4 \log 5 \log 3)$  
A: This is not even true. Just try with n=0
$3^1 = (-5)^{1/2}$ (mod 16) makes no sense.
