Consider a huge $n$ ,such that $n > K^{100000}$ . We claim that the number $x=(n!)^{d-1}$ works . To see this , we prove that $x$ can be written as a $d$ digit palindrome for each $ \frac {n!}{i} -1$ with $1 \leq i < n$ . Indeed we claim that $$ x = \sum_{i=0}^{d-1} k^{d-1} \binom {d-1}{i} \left( \frac {n!}{k} -1 \right)^{i}$$ When we write this in base $ \frac {n!}{k} -1$ ,it is clear that it is a palindrome . Also the fact that $n$ is huge, ensures that all the $k^{d-1} \binom {d-1}{i} < \frac {n!}{k} -1$. Now we prove our claim . This is just a matter of carrying out a summation. We have : $$ \sum_{i=0}^{d-1} k^{d-1} \binom {d-1}{i} \left( \frac {n!}{k} -1 \right)^{i} = k^{d-1} \sum_{i=0}^{d-1} \binom {d-1}{i} \left( \frac {n!}{k} -1 \right)^{i} = k^{d-1} \left( \frac {n!}{k} -1 +1 \right)^{d-1} = k^{d-1} \left( \frac {n!}{k}\right)^{d-1} ={(n!)}^{d-1}$$ We're done $\blacksquare$