10! = 2 * 5 * $3^{2}$ * $2^{3}$ * 7 * (2 * 3) * 5 * $2^{2}$ * 3 * 2 = $2^{8}$ * $3^{4}$ * $5^{2}$ * 7 = ( $2^{4}$ * $3^{2}$ * 5) $^{2}$ * 7

Therefore k = $2^{4}$ * $3^{2}$ * 5 = $\boxed{720}$

It's easy to see why 1 will be the power of 7 (hint: $7 \cdot 2 > 10$ ). So we just find the powers of primes less than 7.

Highest power of 2 -

$\left\lfloor \frac { 10 }{ 2 } \right\rfloor +\left\lfloor \frac { 10 }{ 4 } \right\rfloor +\left\lfloor \frac { 10 }{ 8 } \right\rfloor =8$

Highest power of 3 -

$\left\lfloor \frac { 10 }{ 3 } \right\rfloor +\left\lfloor \frac { 10 }{ 9 } \right\rfloor =4$

Highest power of 5 -

$\left\lfloor \frac { 10 }{ 5 } \right\rfloor =2$

So we have the factorization as ${ 2 }^{ 8 }\cdot { 3 }^{ 4 }\cdot { 5 }^{ 2 }={ \left( { 2 }^{ 4 }\cdot { 3 }^{ 2 }\cdot 5 \right) }^{ 2 }=720^2$

Show $10!$ as a product of it's prime factors.

$10! =10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

$=(5 \times 2) \times (3 \times 3) \times (2 \times 2 \times 2) \times 7 \times (2 \times 3) \times 5 \times (2 \times 2 ) \times 3$

$=2^{8} \times 3^{4} \times 5^{2} \times 7$

$=(2^{4} \times 3^{2} \times 5^{1})^{2} \times 7$

Hence the largest possible $k$ for which $k^{2} | 10!$ is $k=2^{4} \times 3^{2} \times 5 =\boxed{720}$

That's a wonderful and little bit tricky question.But I have found its answer,which is 720 Since 720*720 *7=3628800=10! where "7" being the least factor (integer) whose product with the square of required number (720) gives 10!

$10! = 2^{8} \cdot 5^{2} \cdot 3^{5} \cdot 7 \cdot 1$

$10! = 2^{8} \cdot 5^{2} \cdot 3^{4} \cdot 3 \cdot 7 \cdot 1$

This step has new given us these factors separated into two groups, those with an even exponent and those with an odd exponent. Knowing that our factor will be squared, we collect the even exponent factors, call their product k^{2}, take the square root, ergo dividing all exponents by 2, to yield our desired k

$k^{2} = 2^{8} \cdot 5^{2} \cdot 3^{4}$

$k =2^{4} \cdot 5^{1} \cdot 3^{2}$

$\boxed k = 720$