Let $ f$ be a function with the following properties:
(i) $f(1) \equal{} 1$, and
(ii) $ f(2n) \equal{} n\times f(n)$, for any positive integer $ n$.
What is the value of $ f(2^{100})$?

$ \textbf{(A)}\  1 \qquad \textbf{(B)}\  2^{99} \qquad \textbf{(C)}\  2^{100} \qquad \textbf{(D)}\  2^{4950} \qquad \textbf{(E)}\ 2^{9999}$
To solve for \( f(2^{100}) \), we need to use the given properties of the function \( f \):
1. \( f(1) = 1 \)
2. \( f(2n) = n \times f(n) \) for any positive integer \( n \).

We will start by calculating \( f \) for some small powers of 2 to identify a pattern.

1. \( f(1) = 1 \)
2. \( f(2) = 1 \times f(1) = 1 \)
3. \( f(4) = 2 \times f(2) = 2 \times 1 = 2 \)
4. \( f(8) = 4 \times f(4) = 4 \times 2 = 8 \)
5. \( f(16) = 8 \times f(8) = 8 \times 8 = 64 \)

We observe that:
\[
f(2^1) = 1 = 2^0
\]
\[
f(2^2) = 2 = 2^1
\]
\[
f(2^3) = 8 = 2^3
\]
\[
f(2^4) = 64 = 2^6
\]

We can see a pattern emerging. Let's generalize this pattern. Suppose \( f(2^k) = 2^{\frac{k(k-1)}{2}} \). We will prove this by induction.

**Base Case:**
For \( k = 1 \):
\[
f(2^1) = 1 = 2^0 = 2^{\frac{1(1-1)}{2}}
\]
The base case holds.

**Inductive Step:**
Assume \( f(2^k) = 2^{\frac{k(k-1)}{2}} \) holds for some \( k \geq 1 \). We need to show that \( f(2^{k+1}) = 2^{\frac{(k+1)k}{2}} \).

Using the given property:
\[
f(2^{k+1}) = 2^k \times f(2^k)
\]
By the inductive hypothesis:
\[
f(2^k) = 2^{\frac{k(k-1)}{2}}
\]
Thus:
\[
f(2^{k+1}) = 2^k \times 2^{\frac{k(k-1)}{2}} = 2^{k + \frac{k(k-1)}{2}} = 2^{\frac{k(k+1)}{2}}
\]
The inductive step holds, so by induction, \( f(2^k) = 2^{\frac{k(k-1)}{2}} \) for all \( k \geq 1 \).

Now, we need to find \( f(2^{100}) \):
\[
f(2^{100}) = 2^{\frac{100(100-1)}{2}} = 2^{\frac{100 \times 99}{2}} = 2^{4950}
\]

The final answer is \( \boxed{2^{4950}} \).