Since the product of the odd integers less than $2014$ contains $25$ as a factor, it is clearly divisible by $25$. Also, since it is odd, its last two digits have to be $25$ or $75$. So the product is either of the form $100n + 25$ or of the form $100n + 75$. In either case, the last two digits of the squared product (which is the same as the product of the squares) are $25$:
$$(100n + 25)^2 = 10000n^2 + 5000n + 625 = 100(100n^2 + 50n + 6) + 25$$
or
$$(100n + 75)^2 = 10000n^2 + 15000n + 5625 = 100(100n^2 + 150n + 56) + 25.$$
In fact, we see that the last three digits have to be $625$.

Our number has $25 = 5^2$ as a factor, so its last two digits are $00$, $25$, $50$ or $75$. Moreover, it is the square of an odd number, which we can write as $(2n+1)^2 = 4n^2 + 4n + 1$. This means that the remainder upon division by $4$ has to be $1$. Now note that $100n + 0$, $100n + 25$, $100n + 50$ and $100n + 75$ leave a remainder of $0$, $1$, $2$, $3$ respectively when divided by $4$. This means that the last two digits are in fact $25$.