If $p$ is an odd prime numbers, show that $$\sigma(4^p-1)<(2^{p+1}-1)^2$$ where $\sigma(n)$ stands for the sum of divisors.
I thought of using the formula for $\sigma(n)$:
If $4^p-1=3^k\cdot (p_{1})^{t_{1}}(p_{2})^{t_{2}}\cdots (p_{m})^t_{m}$ we get $$\sigma(4^p-1)=(4^p-1)\left(\prod_{i=0}^{n}\dfrac{1}{3^i}\right)\left(\prod_{1\le i\le m}\left(1+\dfrac{1}{q_{1}}+\cdots+\dfrac{1}{q^{t_{i}}_{i}}\right)\right)$$ but couldn't get any further.