Step 1: Read what is asked.
We want the formal charge sitting on the nitrogen atom inside the ammonium ion $\text{NH}_4^+$, not on the whole ion.
Step 2: Recall the working rule.
Formal charge $= (\text{valence electrons of free atom}) - (\text{lone pair electrons}) - \tfrac{1}{2}(\text{bonding electrons})$. A neat shortcut for a fully bonded atom with no lone pairs is: count the bonds and compare with the normal valency.
Step 3: Set up nitrogen.
A free nitrogen atom has $5$ valence electrons. In $\text{NH}_4^+$ nitrogen makes $4$ N-H single bonds and keeps no lone pair.
Step 4: Count electrons owned by N.
Lone pair electrons $= 0$. Bonding electrons around N $= 4 \times 2 = 8$, so half of them $= 4$.
Step 5: Plug into the formula.
$\text{FC} = 5 - 0 - 4 = +1$. The extra fourth bond means nitrogen has effectively given up a share of one electron, which shows up as a $+1$ charge.
Step 6: Sanity check with the ion charge.
Each H is neutral ($+0$), so the only charge carrier is N. Since the ion as a whole is $+1$, nitrogen must indeed carry $+1$. This confirms option (1).
\[ \boxed{\text{Formal charge on N in } \text{NH}_4^+ = +1} \]