How prove this $n^n+47\equiv 0\pmod{2^l}$ Show that: for any positive integer $l$, there exist positive integers $n$ such
$$n^n+47\equiv 0\pmod{2^l}$$
I feel this problem maybe uses Mathematical induction, but I can't solve it.
Thank you
 A: For $l\leq 4$ we have $n=1$, so suppose $l>4$.
Fisrt, note that the order of the elements of the multiplicative group $\mathbb Z_{2^l}^*$ is at most $\varphi(2^l)=2^{l-1}$, where $\varphi$ is the totient function. In fact, this order can be bounded by $2^{l-2}$ (you can found a simple proof of this in Introduction to Analytic Number Theory by Tom Apostol, p. 206), although this result is not needed for the problem. 
Now, if $n^n+47\equiv0\pmod{2^l}$, but $n^n+47\not\equiv0\pmod{2^{l+1}}$ we have
$$(n+2^l)^{n+2^l}\equiv(n+2^l)^n \pmod{2^{l+1}}$$
and
$$\begin{align}
(n+2^l)^n&=\sum_{j=0}^n\binom nj n^j2^{l(n-j)}=\\
&=n^n+n\cdot n^{n-1}2^l+2^{2l}s=\\
&=n^n(1+2^l)+2^{2l}s
\end{align}$$
for some integer $s$.
Since $2^l$ divides $n^n+47$ but $2^{l+1}$ doesn't, $n^n+47\equiv2^l\pmod{2^{l+1}}$. Taking modulus $2^{l+1}$ in the former equation,
$$\begin{align}(n+2^l)^n &\equiv (2^l-47)(1+2^l)\equiv\\
&\equiv -46\cdot2^l-47\equiv\\
&\equiv-47\pmod{2^{l+1}}
\end{align}$$
q.e.d.
A: Since $\phi(2^l) = 2^{l-1}$ and $2^{l-1}$ divides $2^l$, the value of $n^n$ modulo $2^l$ only depends on $n$ modulo $2^l$. So the question is actually about showing that there is a solution to $n^n = -47$ in the $2$-adic numbers.
Except that exponentiation isn't exactly well-defined.
We can however define $\log : 1 + 4\Bbb Z_2 \to 4\Bbb Z_2$ and $\exp : 4\Bbb Z_2 \to 1+4\Bbb Z_2$ by their respective power series, or directly define $(1+ 4x)^y$ by the power series $\sum \binom y k 4^kx^k$. You end up with a power series for $(1+4x)^{1+4x} = 1+4x+16x^2+32x^3+256x^4/3+256x^5/3 + \ldots$, which shows that $(1+4x) \mapsto (1+4x)^{1+4x}$ is an isometry from $1+4\Bbb Z_2$ to itself. To obtain the solution you can then invert the power series, and apply it to $-47 = 1+4*(-12)$.
