easy to solve

correct

is this supposed to be solved like this by knowing this fact or is there some other way?

1!+2!+3!+4!=33, 5!=120, 6!=720, 7!=5040, 8! = 40320, 9! =326880 i.e the digit in tenth place of 1!+2!+.......+9! is 1 n! is divisible by 100 for all n ≥ 10 n!= 100k_n for all n ≥10, where k₁₀, k₁₁, .........are positive integers.
i.e 10!+11! + .........+49!=100 (k₁₀ +k₁₁ +.......+k₄₉₎

i.e the digit in the tenth place of this sum is 0
i.e the digit in the tenth place of 1!+2!+......+49! is 1.