Prove that there are infinitely many numbers $n$ so that $2^n$ ends in $n$.
I was thinking of using induction. I know that $2^{36}$ ends in $36$ and $2^{736}$ ends in $736$. Indeed, one can see that $2^{100}\equiv 1\pmod{125}$ by the Euler Fermat theorem so $2^{103}\equiv 8\pmod{1000}$. Hence $2^{736} = (2^{103})^{7}\cdot 2^{15} \equiv 8^7\cdot 2^{10}\cdot 2^5\equiv (2^{10})^2\cdot 2\cdot 24\cdot 32\equiv 24^3\cdot 64\equiv 824\cdot 64\equiv 736\pmod{1000}$. Thus the proof will be complete if I can show that if $2^{n}$ ends in $dn$, where $d$ is the digit before $n$, then $2^{dn}$ ends in $dn$, but I'm not sure how to show this inductive step. Perhaps the Euler-Fermat theorem or Fermat's Little theorem might be useful?
Edit: Yes, it seems like this question might be a duplicate but I think that this question needs more justification than the one linked in the first comment below. In particular, I don't see enough justification in the answers for that question as to why certain numbers of the form $n=100l+36$ satisfy $2^n$ ends in $n$.
Best Answer
We prove the stronger result that there is such a number $w_{n}$ in the interval $[2^{n},10^{n}]$ for $n \geq 2$ such that $w_{n}$ is a multiple of $2^n$.
$\textbf{Background}:$ For $n \in \mathbb{N}$ and $p$ a prime we set $\text{ord}_{p}(n)$ to be the unique integer $a \geq 0$ so that $p^{a}|n$ and $p^{a+1} \not \div n$. I.e $\text{ord}_{2}(6) = 1$ and $\text{ord}_{2}(3) = 0.$
$\textbf{Claim}:$ For $n \geq 2$ there exists there exists $v_{n}\in \mathbb{N}$ which satisfies $2^{v_{n}}-v_{n} \equiv 0 \mod 10^{n}$ and $\text{ord}_{2}(v_{n}) \geq n$.
$\textbf{Proof}:$
Set $v_{2} = 36$ so that $\text{ord}_{2}(v_{2}) = 2 \geq 2$. Suppose we have $v_{2},...,v_{k}$ that satisifies the above condition, then
$$\text{ord}_{2}(2^{v_{k}})\geq k+1$$
$$1)\text{ }\text{ }2^{2^{v_{k}}}-2^{v_{k}} \equiv 2^{v_{k}}(2^{2^{v_{k}}-v_{k}}-1)\mod 10^{k+1}$$
Now $$2)\text{ }\text{ }2^{v_{k}} \equiv 0 \mod 2^{k+1}$$
$$3)\text{ }\text{ }\text{ }\begin{cases}2^{v_{k}}-v_{k} \equiv 0 \mod 5^k \\ 2^{v_{k}}-v_{k} \equiv 0 \mod 4\end{cases} \Rightarrow 2^{v_{k}}-v_{k} \equiv 0 \mod 4(5^k) \Rightarrow 2^{2^{v_{k}}-v_{k}}-1 \equiv 0 \mod 5^{k+1}$$
Thus by $1), 2)$ and $3)$
$$2^{2^{v_{k}}}-2^{v_{k}} \equiv 0 \mod 10^{k+1}.$$
Hence it suffices to take $v_{k+1}=2^{v_{k}}$.
$\textbf{Claim}$: For every $n \geq 2$ there exists an integer $d_{n} \in [0,10^{n}-1]$ which satisfies $\text{ord}_{2}(d_{n}) \geq n$ and $2^{d_{n}}-d_{n} \equiv 0 \mod 10^{n}$.
$\textbf{Proof}:$
Since we have $v_{n} = 2^{n}c_{n}$ for some integer $c_{n}$ we know that
$$2^{2^{n}c_{n}}-2^{n}c_{n} \equiv 0 \mod 10^{n}$$
$$2^{2^{n}c_{n}-n}-c_{n} \equiv 0 \mod 5^{n}$$
Note that for all $\gamma \in \mathbb{Z}$
$$2^{2^{n}(c_{n}+\gamma 5^{n})-n}-(c_{n}+\gamma 5^{n}) \equiv 0 \mod 5^{n}$$
There exists a unique $\gamma_{n} \in \mathbb{Z}$ so that $c_{n}+\gamma_{n}5^{n} \in (0,5^{n}-1]$ (it can't be $0$ for trivial reasons).
Now set $d_{n} = 2^{n}(c_{n}+\gamma_{n}5^{n})$.
The sequence $d_{2},d_{3},...$ contains infinitely many distinct non-negative integers such that $2^{d_{j}}$ ends in $d_{j}$ with $d_{j} \in [2^{j},10^{j}]$