1. Simplify the product

We use the identity

$$(1-x)(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})=1-x^{32}.$$

Hence,

$$(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})= \frac{1-x^{32}}{1-x}, \quad x\neq 1.$$

So

$$y(x)=\frac{1-x^{32}}{1-x}=1+x+x^2+\cdots+x^{31}.$$

This polynomial form is valid for all $x$, so we can differentiate termwise.

2. Differentiate

Since

$$y(x)=\sum_{k=0}^{31} x^k,$$

we get

$$y'(x)=\sum_{k=1}^{31} kx^{k-1},$$

and

$$y''(x)=\sum_{k=2}^{31} k(k-1)x^{k-2}.$$

We need $y'(-1)-y''(-1)$.

3. Compute $y'(-1)$

$$y'(-1)=\sum_{k=1}^{31} k(-1)^{k-1}.$$

This is the alternating sum

$$1-2+3-4+\cdots+31.$$

Pair terms:

$$(1-2)+(3-4)+\cdots+(29-30)+31.$$

There are $15$ pairs, each equal to $-1$, so

$$y'(-1)=15(-1)+31=16.$$

4. Compute $y''(-1)$

$$y''(-1)=\sum_{k=2}^{31} k(k-1)(-1)^{k-2}.$$

Since $(-1)^{k-2}=(-1)^k$, this becomes

$$y''(-1)=\sum_{k=2}^{31} k(k-1)(-1)^k.$$

Write terms in pairs:

$$(2\cdot1)-(3\cdot2)+(4\cdot3)-(5\cdot4)+\cdots+(30\cdot29)-(31\cdot30).$$

Now pair as

$$(2\cdot1-3\cdot2)+(4\cdot3-5\cdot4)+\cdots+(30\cdot29-31\cdot30).$$

For a general pair with even $k=2m$:

$$(2m)(2m-1)-(2m+1)(2m)=2m[(2m-1)-(2m+1)]=-4m.$$

Here $m=1,2,\dots,15$. Therefore

$$y''(-1)=\sum_{m=1}^{15}(-4m)=-4\cdot\frac{15\cdot16}{2}=-480.$$

5. Compute the required value

$$y'(-1)-y''(-1)=16-(-480)=496.$$

6. Compare with the stored answer

Derived answer is 496, which matches Option A.