I am given that, k > 0, m >= 1, and x ≡ 1 (mod m^k) And I must show that this implies x^m ≡ (mod m^(k+1))

I have worked out a bit on my own. Since x ≡ 1 (mod m^k), then x = c*m^k + 1
Then I get x^m = (c*m^k + 1)^m
Or m^(k + 1) | (m^k)^2
So (m^k)^b ≡ 0 (mod m^(k + 1) with b <= m

Not sure if that is correct, but it's all I got so far

Let x = 1 (mod m^k).
Then x^m = 1^m = 1 (mod m^k)
=> x^m - 1 = jm^k
=> m(x^m - 1) = jm^(k + 1)
=> mx^m - m = jm^(k + 1)
=> mx^m = m (mod m^(k + 1))
Now, suppose x^m ≠ 1 (mod m^(k + 1)).
Then x^m - 1 ≠ nm^(k + 1).
So mx^m - m ≠ pm^(k + 1).
That is, mx^m ≠ m (mod m^(k + 1).
However, we know that mx^m = m (mod m^(k + 1)).
Hence, x^m = 1 (mod m^(k + 1)).