jmc

Because of the symmetric nature of the number 11, it is a divisor of many palindromes. Expanding powers of $x+y$ (where $x=10$ and $y=1$) helps us see why the first few powers of 11 are all palindromes: $$\begin{array}{rl}(x+y{)}^2& =x^2+2xy+y^2\\ 11^2& =121\\ (x+y{)}^3& =x^3+3x^{2}y+3x{y}^2+y^3\\ 11^3& =1331\\ (x+y{)}^4& =x^4+4x^{3}y+6x^2{y}^2+4x{y}^3+y^4\\ 11^4& =14641\\ (x+y{)}^5& =x^5+5x^{4}y+10x^3{y}^2+10x^2{y}^3+5x{y}^4+y^5\\ 11^5& =161051\end{array}$$ Notice that each term of the form $x^i{y}^{n-i}$ ends up being a power of $10$, and the digits of $(x+y{)}^n$ end up being the binomial coefficients when these coefficients are less than $10$, as there is no carrying. Because of the identity $\left(\genfrac{}{}{0ex}{}{n}{i}\right)=\left(\genfrac{}{}{0ex}{}{n}{n-i}\right)$, the number is a palindrome whenever the coefficients are all less than $10$, which is true for powers less than 5. However, from the list above, we see that $(x+y{)}^5$ has coefficients at least 10, and indeed, we have $11^5=\boxed{161051}$, which is not a palindrome.