For $\exp(-10)$, more terms have to be computed since the 30-th term $\dfrac{(-10)^{30}}{30!}\approx 0.00377$ is several times bigger than the partial sum so far, 0.0009703416.
For $\exp(10)$, all terms are positive. The 30-th term $\dfrac{(10)^{30}}{30!}\approx 0.00377$ is already substantially negligible to the partial sum so far, 22026.464.
Here is the output of the program of my Python program that computes $\exp(-10)$ by that Taylor series. The first column tells the ordinals of the terms. The second columns show the values of that terms. The third terms is the partial sums so far.
0 1.0 1.0
1 -10.0 -9.0
2 50.0 41.0
3 -166.66666666666666 -125.66666666666666
4 416.66666666666663 291.0
5 -833.33333333333326 -542.33333333333326
6 1388.8888888888887 846.55555555555543
7 -1984.1269841269839 -1137.5714285714284
8 2480.1587301587297 1342.5873015873012
9 -2755.7319223985887 -1413.1446208112875
10 2755.7319223985887 1342.5873015873012
11 -2505.2108385441716 -1162.6235369568703
12 2087.6756987868098 925.05216182993945
13 -1605.9043836821616 -680.8522218522221
14 1147.0745597729726 466.22233792075053
15 -764.71637318198179 -298.49403526123126
16 477.94773323873864 179.45369797750737
17 -281.14572543455211 -101.69202745704473
18 156.19206968586229 54.500042228817563
19 -82.206352466243317 -27.706310237425754
20 41.103176233121658 13.396865995695904
21 -19.572941063391266 -6.1760750676953613
22 8.8967913924505755 2.7207163247552142
23 -3.8681701706306848 -1.1474538458754706
24 1.6117375710961186 0.46428372522064798
25 -0.64469502843844739 -0.18041130321779941
26 0.24795962632247978 0.067548323104680369
27 -0.09183689863795548 -0.024288575533275111
28 0.032798892370698378 0.00851031683742327
29 -0.011309962886447721 -0.0027996460490244501
30 0.0037699876288159102 0.00097034157979145996
31 -0.0012161250415535199 -0.00024578346176205997
32 0.00038003907548547002 0.00013425561372341999
33 -0.00011516335620772 1.90922575157e-05
34 3.3871575355210003e-05 5.2963832870910003e-05
35 -9.6775929586300006e-06 4.3286239912280003e-05
36 2.6882202662899999e-06 4.5974460178560002e-05
37 -7.2654601791999997e-07 4.5247914160649999e-05
38 1.9119632050000001e-07 4.5439110481150002e-05
39 -4.9024697569999999e-08 4.539008578359e-05
40 1.225617439e-08 4.5402341957980002e-05
41 -2.9893108300000001e-09 4.5399352647149997e-05
42 7.1174066999999995e-10 4.5400064387830001e-05
43 -1.6552109000000001e-10 4.5399898866740002e-05
44 3.761843e-11 4.539993648517e-05
45 -8.3596500000000006e-12 4.5399928125519999e-05
46 1.81732e-12 4.539992994283e-05
47 -3.8665999999999999e-13 4.5399929556169997e-05
48 8.0549999999999995e-14 4.5399929636719997e-05
49 -1.6440000000000001e-14 4.5399929620280003e-05
50 3.2899999999999998e-15 4.5399929623570001e-05
51 -6.4000000000000005e-16 4.5399929622930003e-05
52 1.2e-16 4.5399929623049997e-05
53 -2.0000000000000001e-17 4.539992962303e-05
54 0.0 4.539992962303e-05
55 -0.0 4.539992962303e-05
4.539992962303116e-05
Although the partial sums becomes stable after 55-th term, not all digits of 4.539992962303e-05 is correct, because of the precision of the terms such as the 8-th term. The actual value of $\exp(-10)$ should be something like 4.5399929762485e-05
In the end, it is faster and easier to compute $\exp(-x)$ to a specified precision using $\dfrac1{\exp(x)}$ instead of directly for $x>0$, especially when $x$ is large.