This isn't a proof, just the method I would use to approach it.
Consider n divides na, and na can be broken down into a prime factorization.
And n divides nb, and nb can be broken down into a prime factorization.
And the highest common factor of two numbers in their prime factorization form is the lowest value of the exponent of each prime factor
So, for example, if a = 20, then it's prime factorization is and if b = 25 then it's prime factorization is and so the highest common factor is
Now, if you multiply both numbers by an integer n, then the prime factorization of n will be in both numbers prime factorization, and will consequently be in the answer for the hcf(na, nb) which means that n divides hcf(na, nb). and n * hcf(a,b) = hcf(na,nb)
So like lets say our last problem we chose n=15, then na=15*20=300, and the prime factorization of na is
and nb = 15*25=375 then it's prime factorization is and so the highest common factor is
and 15 divides 75, leaving 5 leftover.
I hope that made sense, I'm really bad at writing proofs, so I just wanted to give you an example of what method I think would be easiest to show it.