For any m, n positive integers, let P(m;n) the property you want to prove.
P(1;1) is true, since 2 ! / 2² < 1
Notice that P(m;n) is equivalent to P(n;m).
Let m, n positive integers, suppose that P(m;n) is true. Let's prove that P(m;n+1) is also true :
Let A = (m+n+1) ! * m^m * (n+1)^(n+1) / [ (m+n+1)^(m+n+1) * m ! * (n+1) ! ]
I let you check that :
A = B * ( 1 + 1/n )^n * ( 1 + 1 /(m+n) )^(-m-n)
where B = (m+n) ! * m^m * n^n / [ (m+n)^(m+n) * m ! * n ! ]
and we have assumed that B < 1 , ie : P(m;n) is true.
Remember that the sequence ( 1 + 1/k ) ^ k is increasing (and tends to e, as k tends to infinity).
As a consequence, ( 1 + 1/n ) ^ n < ( 1 + 1/(m+n) ) ^ (m+n)
We can conclude that A = B * C, with both B and C < 1, so A < 1.
ie : P(m;n) implies P(m;n+1).
To conclude, P(1;1) implies P(1;n) for every positive integer n.
Since P(1;n) is equivalent to P(n;1), P(n;1) is true for every positive integer n.
And P(n;1) implies P(n;m) for every positive integer m.
Finally, P(m;n) is true for every m,n.