Inspired by Nihar

Let $P(x)=\sum_{k=0}^{26}(27-k)^2x^k$ be the original polynomial.

Then the polynomial with roots $1-a_k$ for $1\le k\le 26$ is $Q(x)=P(1-x)=\sum_{k=0}^{26}(27-k)^2(1-x)^k$

But now we see the constant term of $Q(x)$ is clearly $Q(0)=\sum_{k=0}^{26}(27-k)^2=\sum_{k=1}^{27}k^2=\dfrac{27\cdot 28\cdot 55}{6}=\boxed{6930}$

Let the function be $f(x)$ .

By factor theorem,

$f(x) = (x-a_1)\times(x-a_2)\times(x-a_3)\times\ldots\times(x-a_{26})$

Now the constant of the new monic polynomial be $b_0$ .

Now since it is a monic polynomial, we know that the leading coefficient is $1$ .

Hence $b_0$ is nothing but the product of the roots of the new monic polynomial.

$b_0 = (1-a_1)\times(1-a_2)\times(1-a_3)\times\ldots\times(1-a_{26})$

which from quick observation can be found out to be $f(1)$ .

$f(1) = 1 + 4 + 9 + \ldots + 729 = 1^2 + 2^2 + 3^2 + \ldots + 27^2$

$\Rightarrow f(1) = \dfrac{27\times 28\times 55}{6} = \boxed{6930}$

Note

The last step was achieved by the formula

$\displaystyle \sum_{i=1}^{n} i^2 = \dfrac{(n)(n+1)(2n+1)}{6}$