No calculators? Seriously?

$\large \displaystyle A = \prod_{k=1}^{\infty} k^{\frac{1}{2^{k-1}}}$

$\large \displaystyle \ln(A) = \sum_{k=1}^{\infty} \frac{ \ln(k) }{2^{k-1}}$

$\large \displaystyle \ln(A) = 2 \sum_{k=1}^{\infty} \frac{ \ln(k) }{2^k}$

Since $\ln(1) = 0$ :

$\large \displaystyle \ln(A) = 2 \sum_{k=2}^{\infty} \frac{ \ln(k) }{2^k}$

Since for $k \geq 3$ we have $\ln(k) > 1$ and, for although $\ln(2) < 1$ , in the sum all the terms for $3$ and above make:

$\large \displaystyle \ln(A) > 2 \sum_{k=2}^{\infty} \frac{ 1 }{2^k}$

$\large \displaystyle \ln(A) > 2 \left ( \frac{\frac14}{1 - \frac12} \right )$

$\large \displaystyle \ln(A) > 1$

$\large \displaystyle A > e = B$

$\color{#3D99F6} \boxed{ \large \displaystyle A > B}$