Prove that $11 | 10^{2n+1}+1$ for all $n\in \mathbb{N}\cup \{0\}$. $$11 | 10^{2n+1}+1\;\;\; \forall n\in \mathbb{N}\cup\{0\}  \tag{$\star$}$$
My proof of $(\star)$ is as follows:
\begin{align}
10^{2n+1}+1
&= 10\cdot10^{2n}+1 \\
&= (11-1)\cdot10^{2n}+1 \\
&= 11\cdot10^{2n}-10^{2n}+1 \\
&= 11\cdot10^{2n}-\left(10^{2n}-1\right) \\
&= 11\cdot10^{2n}-\left(100^{n}-1\right) \\
&= 11\cdot10^{2n}-\left((99+1)^{n}-1\right) \\
&= 11\cdot10^{2n}-\left(1+\binom{n}{1}99+\binom{n}{2}99^2+\cdots+\binom{n}{n-1}99^{n-1}+99^n-1\right) \\
&= 11\cdot10^{2n}-\underbrace{99}_{11\cdot9} \left(\binom{n}{1}+\binom{n}{2}99+\cdots+\binom{n}{n-1}99^{n-2}+99^{n-1}\right) \\
&= 11\left(10^{2n}-9 \left(\binom{n}{1}+\binom{n}{2}99+\cdots+\binom{n}{n-1}99^{n-2}+99^{n-1}\right)\right)
\end{align}
Is there an easier way to prove $(\star)$? The expansion of $(99+1)^n$ seems unnecessarily complicated, but I wasn't sure how else to go from there. Easier proofs are welcome!
 A: Use modular arithmetic:
$$10\equiv-1\mod11$$
$$10^2\equiv1\mod11$$
$$10^{2n}\equiv1\mod11$$
$$10^{2n+1}\equiv-1\mod11$$
$$11|10^{2n+1}+1$$
A: Because for $n\geq1$ we have: $$10^{2n+1}+1=(10+1)(10^{2n}-10^{2n-1}+...-10+1).$$
For $n=0$ it's obvious.
A: $2n+1$ is odd and $10\cong -1\pmod {11}$. Thus $10^{2n+1}\cong (-1)^{2n+1}\cong -1\pmod{11}$.
A: Not as easy as the other answers, but this is a classic induction question.
For $n = 0$, we have $10^{2n + 1} + 1 = 11$, which $11$ divides, confirming the base case.
Suppose $11$ divides $10^{2n + 1} + 1$ for some $n$. Then some $k$ exists such that $10^{2n + 1} + 1 = 11k$. Then,
$$10^{2(n + 1) + 1} + 1 = 100 \cdot 10^{2n + 1} + 1 = 100 (11k - 1) + 1 = 11(100k - 9),$$
thus $11$ divides $10^{2(n+1) + 1} + 1$. By induction, $11$ divides $10^{2n + 1} + 1$ for all $n$.
A: Use the identity
$a^{2n+1}+b^{2n+1}
=(a+b)\sum_{j=0}^{2n} (-1)^j a^j b^{2n-j}
$.
Then put
$a=10, b=1$.
Proof.
$\begin{array}\\
(a+b)\sum_{j=0}^{2n} (-1)^j a^j b^{2n-j}
&=a\sum_{j=0}^{2n} (-1)^j a^j b^{2n-j}+b\sum_{j=0}^{2n} (-1)^j a^j b^{2n-j}\\
&=\sum_{j=0}^{2n} (-1)^j a^{j+1} b^{2n-j}+\sum_{j=0}^{2n} (-1)^j a^j b^{2n-j+1}\\
&=\sum_{j=1}^{2n+1} (-1)^{j-1} a^{j} b^{2n-j+1}+\sum_{j=0}^{2n} (-1)^j a^j b^{2n-j+1}\\
&=\sum_{j=1}^{2n} (-1)^{j-1} a^{j} b^{2n-j+1}+(-1)^{2n} a^{2n+1} b^{0}+\sum_{j=1}^{2n} (-1)^j a^j b^{2n-j+1}+(-1)^0 a^0 b^{2n+1}\\
&=\sum_{j=1}^{2n} ((-1)^{j-1}+(-1)^j) a^{j} b^{2n-j+1}+a^{2n+1}+ b^{2n+1}\\
&=a^{2n+1}+ b^{2n+1}\\
\end{array}
$
A: $$10^{2n+1}+1 \equiv (-1)^{2n+1} + 1 \equiv -1 + 1 \equiv 0 \pmod{11}.$$
