I'm reading "A classical intro to modern number theory" by Ireland & Rosen, in which Proposition 4.2.4 proof is left as an exercise.
Let $n=2^lk $ with $k $ odd, assuming that $x^n\equiv a (2^e) $ is solvable for $e\geq 2l+1$, then it's solvable for any $e'\geq e $.
Part of the proof is assuming $x_0^n\equiv a (2^e) $ and then to prove that the equation is solvable for $e+1$. We define:
$x_1 = x_0+b2^{e-l}$, then we raise by $n $ and equate to $a (2^{e+1} )$ and seek to get a solvable linear equation for $b$. This means: $a \equiv x_1^n = x_0^n+kx_0^{n-1}b2^e+k(n-1)x_0^{n-2}b^22^{2e-2l+l-1}+\ldots (2^{e+1} $. We wish to be left only with linear elements of $b $, so we restrict $2e-2l+l-1 \geq e+1$ or $e\geq l+2$.
We've achieved the constraint $e\geq l+2$ rather than $e\geq 2l+1$, and I wonder where did we go wrong.
Thanks a lot.