I know that it can be proved by the divisible algorithm but how to proceed as if $a!b!{\,|\,}(a+b-1)!$ so this means $a!b!{\,|\,}a$ and $a!b!{\,|\,}b$ and also it will divide $1$ so this means $a!b!=1$ otherwise it will not divide $1$.
Is my proceed is in right direction?
I presume you want to prove $$a!b!\mid(a,b)(a+b-1)!\tag{1}$$ where $(a,b)=\gcd(a,b)$. Using Bezout's identity, this will follow if we can prove both $$a!b!\mid a(a+b-1)!\tag{2}$$ and $$a!b!\mid b(a+b-1)!.\tag{3}$$ These are similar, so just consider $(2)$. This is equivalent to $$(a-1)!b!\mid (a+b-1)!$$ which is true, as the binomial coefficient $\binom{a+b-1}{a-1}$ is an integer.