Does entangling 100 qubits require a 2^100 X 2^100 Hadamard gate?
It does require that, but you get it for free.

When you apply a 2x2 gate to a single qubit, the gate you are applying to the $n$-qubit system that qubit is in is automatically $2^n$ x $2^n$. If you apply an $H$ to the third qubit out of six, then the operation you applied to the six qubits is $I_2 \otimes I_2 \otimes H \otimes I_2 \otimes I_2 \otimes I_2$. That's a 64x64 matrix; exactly what you needed.

So performing a Hadamard transform on $n$ qubits is actually very easy, even though some very large matrices are being thrown around if you think about what's happening in that way. Just independently hit each qubit with the usual single-qubit $H$ gate.

Note that Hadamarding each qubit doesn't entangle them, though. It puts them in a superposition, but the state still factors. If you want to entangled all the qubits, you could instead hit one of them with a Hadamard then use Controlled-Nots to toggle each of the other qubits based on that Hadamard'ed qubit.