I have started this proof with the base case $n = 10$. In this case, it's true because $1000 < 1024$.
Next I assume that $n \geq 11$ and that $100(n-1) \leq {2}^{n-1}$. As the right side is already bigger, we can multiply it by 2 without changing the inequality, hence we have that $100(n-1) \leq 2^n$. But I am currently stuck in this part. How would I proceed from here? I know I need to show that $100n \leq 2^n$.
Once you have shown $P(n)$ to be true for your base case, there is no need to perform any further instantiation. Instead you state the induction hypothesis: $P(n)$ is true for some $n=k$ in the domain. Then perform the induction step and proceed normally. Let $n=k+1$; then $100(k+1) = 100k+100$. By the induction hypothesis, $100k \le 2^k$, which means that $100k+100 \le 2^k+100$. $100 \le 2^k$, therefore $100k+100 \le 2^k+2^k = 2×2^k = 2^{k+1}$. QED.