To solve for $x$
$2^x \equiv 41 \mod 211$
$\phi(p) = p - 1 = 210 = 2.3.5.7$
Solving by Pohlig-Hellman, we get to
$x \equiv 3 \mod 7$
$x \equiv 2 \mod 5$
$x \equiv 2 \mod 3$
$x \equiv 1 \mod 2$
Then using Chinese Remainder theorem we get
$x \equiv 17 \mod 210$
I understood everything up to the above step.
But 17 is the value of $x$ for modulo 210. How do we go from here to $2^{17} \equiv 41 \mod 211$?
What is the relation between $x$ for modulo 210 & the $2^x$ in modulo 211? We already knew $x$ for 2, 3, 5 & 7. Why don't any of those values work as x for $2^x$ for modulo 211?
$2^{x\equiv 17\bmod 210}=2^{17}\times (2^{210})^k$
$ 2^{210}\equiv 1 \bmod 211$
$2^x\equiv 2^{17} \bmod 211\equiv 41\bmod 211$