1. We need the last two digits of $1919^{1919}$, i.e. the value modulo $100$.

2. Reduce the base modulo $100$: $1919 \equiv 19 \pmod{100}$ So we need $19^{1919} \pmod{100}.$

3. Find a pattern in powers of $19$ modulo $100$:

$19^1 = 19 \equiv 19 \pmod{100}$ $19^2 = 361 \equiv 61 \pmod{100}$ $19^3 = 19\cdot 61 = 1159 \equiv 59 \pmod{100}$ $19^4 = 19\cdot 59 = 1121 \equiv 21 \pmod{100}$ $19^5 = 19\cdot 21 = 399 \equiv 99 \pmod{100}$ $19^6 = 19\cdot 99 = 1881 \equiv 81 \pmod{100}$ $19^7 = 19\cdot 81 = 1539 \equiv 39 \pmod{100}$ $19^8 = 19\cdot 39 = 741 \equiv 41 \pmod{100}$ $19^9 = 19\cdot 41 = 779 \equiv 79 \pmod{100}$ $19^{10} = 19\cdot 79 = 1501 \equiv 1 \pmod{100}$

Thus the powers repeat with period $10$: $19^{10} \equiv 1 \pmod{100}.$

4. Reduce the exponent modulo $10$: $1919 \equiv 9 \pmod{10}$ Hence $19^{1919} \equiv 19^9 \pmod{100}.$ From above, $19^9 \equiv 79 \pmod{100}.$

So the last two digits are $79$.

5. Their product is $7\times 9 = 63.$

Therefore, the required integer is $63$.