Simplify the Expression: m^(-n) × n^(-m) × 1/m

$$We will now apply the law of negative exponents, but in reverse:

$\frac{1}{a^x} =a^{-x}$

We'll apply this law to the problem for the third term in the product:

$m^{-n}\cdot n^{-m}\cdot\frac{1}{m}=m^{-n}\cdot n^{-m}\cdot m^{-1}=m^{-n}\cdot m^{-1}\cdot n^{-m}$

When in the first stage we applied the above law of exponents for the third term in the product, and in the next stage we rearranged the resulting expression using the distributive property of multiplication so that terms with identical bases are adjacent to each other,

Next, we'll recall the law of exponents for multiplying terms with identical bases:

$a^x\cdot a^y=a^{x+y}$

And we'll apply this law of exponents to the expression that we obtained in the last stage:

$m^{-n}\cdot m^{-1}\cdot n^{-m}=m^{-n+(-1)}\cdot n^{-m}=m^{-n-1}\cdot n^{-m}$

When in the first stage we applied the above law of exponents for terms with identical bases, and in the next stage we simplified the expression with the exponent of the first term in the product in the expression we obtained the following,

Let's summarize the solution so far as shown below:

$m^{-n}\cdot n^{-m}\cdot\frac{1}{m}=m^{-n}\cdot n^{-m}\cdot m^{-1}=m^{-n-1}\cdot n^{-m}$

Now let's note that there is no such answer among the given options, and an additional check of what we've done so far will reveal that there is no calculation error,

Therefore, we can conclude that additional mathematical manipulation is needed to determine which is the correct answer among the given options,

Let's note that options B and D have expressions similar to the expression we got in the last stage, while the other two options can be directly eliminated since they are clearly different from the expression we got,

Furthermore, let's note that in addition, the second term in the product in the expression we got, which is the term-$n^{-m}$, is in the numerator (note at the end of the solution on this topic), while in option B it's in the denominator, so we'll eliminate this option,

Thus - we're left with only one option - which is answer D, however we want to verify (and must verify!) that this is indeed the correct answer:

We'll do this using the law of exponents for negative exponents that we mentioned earlier, but in the forward direction:

$a^{-x} = \frac{1}{a^x}$

And we'll deal separately with the first term in the product in the expression we got in the last stage of solving the problem, which is the term:

$m^{-n-1}$

Let's note that we can represent the expression in the exponent as follows:

$-n-1=-(n+1)$

Where we used factoring out and took out negative one from the parentheses,

Next, we'll use the above law of exponents and the last understanding to represent the above expression (which we're currently dealing with, separately) as a term in the denominator of a fraction:

$m^{-n-1}=m^{-\underline {\bm{(n+1)}}}=\frac{1}{m^{\underline {\bm{n+1}}}}$

When in the first stage, in order to use the above law of exponents - we represented the term in question as having a negative exponent, while using the fact that:

$-n-1=-(n+1)$,

Next, we applied the above law of exponents carefully, since the number that- x represents in our use of the above law of exponents here is:

$n+1$(underlined in the expression above)

Let's return then to the expression we got in the last stage of solving the given problem, and apply for the first term in the product the mathematical manipulation we just performed:

$m^{-n-1}\cdot n^{-m}=m^{-(n+1)}\cdot n^{-m}=\frac{1}{m^{n+1}}\cdot n^{-m}$

Now let's simplify the expression that we obtained and perform the multiplication in the fraction while remembering that multiplication in a fraction means multiplying the numerators:

$\frac{1}{m^{n+1}}\cdot n^{-m}=\frac{1\cdot n^{-m}}{m^{n+1}}=\frac{n^{-m}}{m^{n+1}}$

Let's summarize then the solution stages so far as follows:

$m^{-n}\cdot n^{-m}\cdot\frac{1}{m}=m^{-n-1}\cdot n^{-m} =\frac{n^{-m}}{m^{n+1}}$

Therefore, the correct answer is indeed answer D.

Note:

When it's written "the number in the numerator" even though there isn't actually a fraction in the expression, this is because we can always refer to any number as being in the numerator of a fraction if we remember that any number divided by 1 equals itself, meaning, we can always write a number as a fraction like this:

$X=\frac{X}{1}$and therefore we can actually refer to $X$as a number in the numerator of a fraction.